Fundamental integration Questions in English

(A) Let $I = \int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} d x$
We can rewrite the expression inside the square root as a perfect square:
$x^4 + x^{-4} + 2 = (x^2)^2 + (x^{-2})^2 + 2(x^2)(x^{-2}) = (x^2 + x^{-2})^2$
Substituting this into the integral:
$I = \int \frac{\sqrt{(x^2 + x^{-2})^2}}{x^3} d x = \int \frac{x^2 + x^{-2}}{x^3} d x$
Now,divide each term in the numerator by $x^3$:
$I = \int \left( \frac{x^2}{x^3} + \frac{x^{-2}}{x^3} \right) d x = \int \left( \frac{1}{x} + x^{-5} \right) d x$
Integrating term by term:
$I = \int \frac{1}{x} d x + \int x^{-5} d x = \ln |x| + \frac{x^{-4}}{-4} + C$
$I = \ln |x| - \frac{1}{4x^4} + C$

(A) We know that the Taylor series expansion for the exponential function $e^x$ is given by:
$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$
Substituting this into the integral,we get:
$\int \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots \infty \right) dx = \int e^x dx$
The integral of $e^x$ with respect to $x$ is $e^x + c$,where $c$ is the constant of integration.
Therefore,the final answer is $e^x + c$.

(C) Given the integral $I = \int \frac{1+\cos (4 x)}{\cot x-\tan x} dx$.
Using the identity $1+\cos(2\theta) = 2\cos^2(\theta)$,we have $1+\cos(4x) = 2\cos^2(2x)$.
Also,$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos(2x)}{\frac{1}{2}\sin(2x)} = 2\cot(2x)$.
Substituting these into the integral:
$I = \int \frac{2\cos^2(2x)}{2\cot(2x)} dx = \int \frac{\cos^2(2x)}{\frac{\cos(2x)}{\sin(2x)}} dx = \int \cos(2x)\sin(2x) dx$.
Using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$,we have $\sin(4x) = 2\sin(2x)\cos(2x)$,so $\sin(2x)\cos(2x) = \frac{1}{2}\sin(4x)$.
$I = \int \frac{1}{2}\sin(4x) dx = \frac{1}{2} \left( \frac{-\cos(4x)}{4} \right) + c = -\frac{1}{8}\cos(4x) + c$.
Comparing this with $k\cos(4x) + c$,we get $k = -\frac{1}{8}$.
Thus,option $(c)$ is correct.

(B) Given,$\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx$
$= \frac{1}{2} \int (2 \cos 5 x \cos x) \cos 2 x \, dx$
$= \frac{1}{2} \int (\cos 6 x + \cos 4 x) \cos 2 x \, dx$
$= \frac{1}{4} \int (2 \cos 6 x \cos 2 x + 2 \cos 4 x \cos 2 x) \, dx$
$= \frac{1}{4} \int (\cos 8 x + \cos 4 x + \cos 6 x + \cos 2 x) \, dx$
$= \frac{1}{4} \left[ \frac{\sin 8 x}{8} + \frac{\sin 4 x}{4} + \frac{\sin 6 x}{6} + \frac{\sin 2 x}{2} \right] + k$
$= \frac{\sin 2 x}{8} + \frac{\sin 4 x}{16} + \frac{\sin 6 x}{24} + \frac{\sin 8 x}{32} + k$
Comparing with $A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$,we get:
$A = \frac{1}{8}, B = \frac{1}{16}, C = \frac{1}{24}, D = \frac{1}{32}$
Then,$\frac{1}{B} + \frac{1}{C} = 16 + 24 = 40$
And $\frac{1}{A} + \frac{1}{D} = 8 + 32 = 40$
Therefore,$\frac{1}{B} + \frac{1}{C} = \frac{1}{A} + \frac{1}{D}$.

(C) To solve the integral $\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x$,we first observe that the degree of the numerator $(8)$ is greater than the degree of the denominator $(4)$.
Performing polynomial long division of $x^8-9 x^2+18$ by $x^4-3 x^2+3$:
$x^8-9 x^2+18 = (x^4-3 x^2+3)(x^4+3 x^2+6) + 0$.
Thus,the integral becomes:
$\int (x^4+3 x^2+6) d x$.
Integrating term by term with respect to $x$:
$= \int x^4 d x + 3 \int x^2 d x + 6 \int 1 d x$.
$= \frac{x^5}{5} + 3 \left( \frac{x^3}{3} \right) + 6x + c$.
$= \frac{x^5}{5} + x^3 + 6x + c$.

(B) We have the integral $I = \int \frac{dx}{1+\sin x}$.
Multiply the numerator and denominator by $(1-\sin x)$:
$I = \int \frac{1-\sin x}{1-\sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx$
$I = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$
Using half-angle identities $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ and $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2})$:
$I = \frac{\sin x - 1}{\cos x} + C = \frac{-(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} + C$
$I = -\frac{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}{\cos(\frac{x}{2}) + \sin(\frac{x}{2})} + C = -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) + C$
Since $-\tan(A) = \tan(-A)$,we have $I = \tan\left(\frac{x}{2} - \frac{\pi}{4}\right) + C$.
Comparing this with $\tan(\frac{x}{2} - \theta) + C$,we get $\theta = \frac{\pi}{4}$.

(C) Given the integral: $I = \int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$
We can rewrite the terms as: $8^{1+x} = 8 \cdot 8^x = 8 \cdot (2^3)^x = 8 \cdot 2^{3x}$ and $4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x}$
Substituting these into the integral: $I = \int \frac{8 \cdot 2^{3x} + 4 \cdot 2^{2x}}{2^{2x}} dx$
Dividing each term by $2^{2x}$: $I = \int (8 \cdot 2^{3x-2x} + 4) dx = \int (8 \cdot 2^x + 4) dx$
Integrating with respect to $x$: $I = 8 \int 2^x dx + \int 4 dx$
Using the formula $\int a^x dx = \frac{a^x}{\log_e a} + C$: $I = 8 \cdot \frac{2^x}{\log_e 2} + 4x + C$

(D) For statement $(A)$: The left side is $\int_a^b \frac{d}{d x}(f(x)) d x = f(b) - f(a)$,which is a constant value. The right side is $\frac{d}{d x} \int_a^b f(x) d x$. Since $\int_a^b f(x) d x$ is a definite integral resulting in a constant,its derivative with respect to $x$ is $0$. Thus,$f(b) - f(a) \neq 0$ in general,so $(A)$ is false.
For statement $(B)$: By the definition of the indefinite integral,$\frac{d}{d x} \int f(x) d x = f(x)$. The constant $C$ is only present in the result of the indefinite integral itself,not in its derivative. Therefore,$\frac{d}{d x} \left( \int f(x) d x \right) = f(x)$,not $f(x) + C$. Thus,$(B)$ is false.
Conclusion: Both $(A)$ and $(B)$ are false.

(B) Given that $f^{\prime}(x)=x+\frac{1}{x}$.
To find $f(x)$,we integrate both sides with respect to $x$:
$f(x) = \int f^{\prime}(x) dx = \int \left(x + \frac{1}{x}\right) dx$.
Using the linearity property of integration:
$f(x) = \int x dx + \int \frac{1}{x} dx$.
Applying the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1}$ and $\int \frac{1}{x} dx = \log |x| + C$:
$f(x) = \frac{x^2}{2} + \log |x| + C$.

(A) Step $1$: Analyze the Reason $(R)$. We know that $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$ for $u \in [-1, 1]$. Thus,the Reason $(R)$ is true.
Step $2$: Simplify the integrand in Assertion $(A)$. Using the identity from $(R)$,$\sin^{-1}(\log x) + \cos^{-1}(\log x) = \frac{\pi}{2}$ provided $|log x| \le 1$,i.e.,$x \in [1/e, e]$.
Step $3$: Evaluate the integral: $\int \sqrt{x-3} \cdot \frac{\pi}{2} dx = \frac{\pi}{2} \int (x-3)^{1/2} dx$.
Step $4$: Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1}$,we get $\frac{\pi}{2} \cdot \frac{(x-3)^{3/2}}{3/2} + c = \frac{\pi}{2} \cdot \frac{2}{3}(x-3)^{3/2} + c = \frac{\pi}{3}(x-3)^{3/2} + c$.
Step $5$: Since the result matches the Assertion $(A)$,$(A)$ is true and $(R)$ is the correct explanation.

(D) Let $I = \int \left(x \sqrt{x} - e^{\log(\sec x \tan x)} + \frac{3x^2 - 2x + 1}{x^2}\right) dx$.
Using the property $e^{\log f(x)} = f(x)$,we get:
$I = \int \left(x^{3/2} - \sec x \tan x + 3 - \frac{2}{x} + \frac{1}{x^2}\right) dx$.
Integrating term by term:
$\int x^{3/2} dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^2 \sqrt{x}$.
$\int -\sec x \tan x dx = -\sec x$.
$\int 3 dx = 3x$.
$\int -\frac{2}{x} dx = -2 \log x$.
$\int \frac{1}{x^2} dx = \int x^{-2} dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$.
Combining these,we get $I = \frac{2}{5} x^2 \sqrt{x} - \sec x + 3x - 2 \log x - \frac{1}{x} + c$.

(C) Given: $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$
We simplify the integrand: $f(x)=\int\left(2 x-3+\frac{4}{x}-\frac{5}{x^2}\right) d x$
Integrating term by term: $f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}+C$
Given $f(1)=1$,we substitute $x=1$: $1^2-3(1)+4 \log(1)+\frac{5}{1}+C=1$
$1-3+0+5+C=1 \Rightarrow 3+C=1 \Rightarrow C=-2$
Thus,$f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}-2$
Now,calculate $f(5)$: $f(5)=5^2-3(5)+4 \log 5+\frac{5}{5}-2$
$f(5)=25-15+4 \log 5+1-2$
$f(5)=9+4 \log 5$

(C) Given the integral $I = \int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx$.
Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$,we have $\sqrt{1 + \sin 2x} = |\sin x + \cos x|$.
Given $\frac{3 \pi}{4} < x < \frac{7 \pi}{4}$,in this interval,$\sin x + \cos x$ is negative.
Thus,$|\sin x + \cos x| = -(\sin x + \cos x)$.
Substituting this into the integral:
$I = \int \left(2^x - (-(\sin x + \cos x)) + \frac{1}{x^2} - \frac{1}{x}\right) dx$
$I = \int \left(2^x + \sin x + \cos x + \frac{1}{x^2} - \frac{1}{x}\right) dx$
Integrating term by term:
$I = \frac{2^x}{\log 2} - \cos x + \sin x - \frac{1}{x} - \log |x| + c$
Rearranging the terms,we get:
$I = \frac{2^x}{\log 2} + \sin x - \cos x - \frac{1}{x} - \log |x| + c$.
Therefore,option $C$ is correct.

(D) Given the integral $I = \int \left( \frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} \right)^2 dx$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = 2 \cos^2 x - 1$:
$I = \int \left( \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + 2 \cos^2 x - 1} \right)^2 dx$
$I = \int \left( \frac{\sin x (1 + 2 \cos x)}{\cos x (1 + 2 \cos x)} \right)^2 dx$
Since $\cos x \neq -1/2$,we can cancel $(1 + 2 \cos x)$:
$I = \int \left( \frac{\sin x}{\cos x} \right)^2 dx = \int \tan^2 x dx$
Using the identity $\tan^2 x = \sec^2 x - 1$:
$I = \int (\sec^2 x - 1) dx = \tan x - x + c$.

(A) Given $f(x) = \int \frac{2-3 \sin^2 x}{1+\cos 2x} dx$.
Using the identity $1+\cos 2x = 2 \cos^2 x$,we have:
$f(x) = \int \frac{2-3 \sin^2 x}{2 \cos^2 x} dx = \int \left( \frac{2}{2 \cos^2 x} - \frac{3 \sin^2 x}{2 \cos^2 x} \right) dx$
$f(x) = \int (\sec^2 x - \frac{3}{2} \tan^2 x) dx$
Since $\tan^2 x = \sec^2 x - 1$,we get:
$f(x) = \int (\sec^2 x - \frac{3}{2}(\sec^2 x - 1)) dx = \int (\sec^2 x - \frac{3}{2} \sec^2 x + \frac{3}{2}) dx$
$f(x) = \int (\frac{3}{2} - \frac{1}{2} \sec^2 x) dx = \frac{3}{2} x - \frac{1}{2} \tan x + C$
Given $f\left(\frac{\pi}{4}\right) = 1$:
$\frac{3}{2} \left(\frac{\pi}{4}\right) - \frac{1}{2} \tan\left(\frac{\pi}{4}\right) + C = 1$
$\frac{3\pi}{8} - \frac{1}{2}(1) + C = 1 \Rightarrow C = 1 + \frac{1}{2} - \frac{3\pi}{8} = \frac{3}{2} - \frac{3\pi}{8} = \frac{12-3\pi}{8} = \frac{3}{8}(4-\pi)$
Now,$f(0) = \frac{3}{2}(0) - \frac{1}{2} \tan(0) + C = C = \frac{3}{8}(4-\pi)$.

(A) Given the integral $I = \int(\sqrt{1+\sin 2 x} + \sqrt{1-\sin 2 x}) \, dx$.
Using the identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:
$\sqrt{1+\sin 2x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$
$\sqrt{1-\sin 2x} = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$
For $x \in \left(\frac{3 \pi}{4}, \pi\right)$,$\sin x > 0$ and $\cos x < 0$. Also,$|\sin x| > |\cos x|$,so $\sin x + \cos x > 0$ and $\sin x - \cos x > 0$.
Thus,the expression becomes $(\sin x + \cos x) + (\sin x - \cos x) = 2 \sin x$.
Therefore,$I = \int 2 \sin x \, dx = -2 \cos x + C$.

(A) Given integral $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$.
Using the identity $1+\cot^2 x = \operatorname{cosec}^2 x$,we expand the expression inside the square root:
$I = \int \sqrt{1+2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$.
Since $1+\cot^2 x = \operatorname{cosec}^2 x$,we have $1 = \operatorname{cosec}^2 x - \cot^2 x$.
Substituting this:
$I = \int \sqrt{\operatorname{cosec}^2 x - \cot^2 x + 2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$
$I = \int \sqrt{\operatorname{cosec}^2 x + \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$
$I = \int \sqrt{(\operatorname{cosec} x + \cot x)^2} \, dx$
$I = \int (\operatorname{cosec} x + \cot x) \, dx$
Integrating term by term:
$I = \ln |\operatorname{cosec} x - \cot x| + \ln |\sin x| + c$
$I = \ln |(\operatorname{cosec} x - \cot x) \sin x| + c$
$I = \ln |1 - \cos x| + c$
Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$:
$I = \ln |2 \sin^2 \frac{x}{2}| + c = \ln 2 + 2 \ln |\sin \frac{x}{2}| + c$
Since $\ln 2$ is a constant,we can absorb it into $c$:
$I = 2 \ln |\sin \frac{x}{2}| + c$.

(A) We have the integral $I = \int(\cot x \cot (x+\alpha)+1) d x$.
Using the identity $\cot A \cot B + 1 = \frac{\cos A \cos B + \sin A \sin B}{\sin A \sin B} = \frac{\cos(A-B)}{\sin A \sin B}$,we get:
$I = \int \frac{\cos(x - (x+\alpha))}{\sin x \sin (x+\alpha)} d x = \int \frac{\cos(-\alpha)}{\sin x \sin (x+\alpha)} d x = \cos \alpha \int \frac{1}{\sin x \sin (x+\alpha)} d x$.
Multiply and divide by $\sin \alpha$:
$I = \frac{\cos \alpha}{\sin \alpha} \int \frac{\sin \alpha}{\sin x \sin (x+\alpha)} d x = \cot \alpha \int \frac{\sin((x+\alpha)-x)}{\sin x \sin (x+\alpha)} d x$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \cot \alpha \int \frac{\sin(x+\alpha) \cos x - \cos(x+\alpha) \sin x}{\sin x \sin (x+\alpha)} d x$.
$I = \cot \alpha \int (\cot x - \cot (x+\alpha)) d x$.
Integrating,we get:
$I = \cot \alpha (\log |\sin x| - \log |\sin (x+\alpha)|) + c = \cot \alpha \log \left|\frac{\sin x}{\sin (x+\alpha)}\right| + c$.

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-\tan x} dx$.
Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x$.
Substituting these into the integral:
$I = \int \frac{2\cos^2 2x}{2\cot 2x} dx = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} dx = \int \sin 2x \cos 2x dx$.
Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$:
$I = \int \frac{1}{2} \sin 4x dx = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + C = -\frac{1}{8} \cos 4x + C$.

(B) We have the integral $I = \int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx$.
Split the integral into two parts: $I = \int \frac{1}{x^2} dx + \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} dx + \int \frac{\cos^3 x}{\sin^2 x \cos^2 x} dx$.
Simplify each term: $I = \int x^{-2} dx + \int \frac{\sin x}{\cos^2 x} dx + \int \frac{\cos x}{\sin^2 x} dx$.
Evaluate the integrals: $\int x^{-2} dx = -\frac{1}{x}$.
For the second term,let $u = \cos x$,then $du = -\sin x dx$,so $\int \sec x \tan x dx = \sec x$.
For the third term,let $v = \sin x$,then $dv = \cos x dx$,so $\int \csc x \cot x dx = -\csc x$.
Combining these,we get $I = -\frac{1}{x} + \sec x - \csc x + c$.
This can be written as $-\frac{1}{x} + \frac{1}{\cos x} - \frac{1}{\sin x} + c = -\frac{1}{x} + \frac{\sin x - \cos x}{\sin x \cos x} + c$.

(D) Let $u = x+a$,then $du = dx$ and $x = u-a$.
Substituting these into the integral:
$\int \frac{u-a}{u^5} du = \int (u^{-4} - au^{-5}) du$
$= \frac{u^{-3}}{-3} - a \frac{u^{-4}}{-4} + C$
$= -\frac{1}{3u^3} + \frac{a}{4u^4} + C$
$= \frac{-4u + 3a}{12u^4} + C$
$= \frac{-4(x+a) + 3a}{12(x+a)^4} + C$
$= \frac{-4x - 4a + 3a}{12(x+a)^4} + C$
$= \frac{1}{12(x+a)^4}(-4x - a) + C$
Comparing this with $\frac{1}{k(a+x)^4}(f(x)) + c$,we get $k = 12$ and $f(x) = -4x - a$.
Now,calculate $f(-a) = -4(-a) - a = 4a - a = 3a$.
Finally,$\frac{f(-a)}{ak} = \frac{3a}{12a} = \frac{3}{12} = \frac{1}{4}$.

(C) Given $f\left(\frac{2x+1}{5x+3}\right) = x+2$.
Let $t = \frac{2x+1}{5x+3}$.
Then $t(5x+3) = 2x+1 \implies 5xt + 3t = 2x+1 \implies x(5t-2) = 1-3t \implies x = \frac{1-3t}{5t-2} = \frac{3t-1}{2-5t}$.
Substituting $x$ into the function: $f(t) = \frac{3t-1}{2-5t} + 2 = \frac{3t-1 + 4-10t}{2-5t} = \frac{3-7t}{2-5t} = \frac{7t-3}{5t-2}$.
Now,$\int f(x) dx = \int \frac{7x-3}{5x-2} dx$.
Using division: $\frac{7x-3}{5x-2} = \frac{7}{5} \left(\frac{5x-2}{5x-2}\right) + \frac{\frac{14}{5}-3}{5x-2} = \frac{7}{5} - \frac{1}{5(5x-2)}$.
Thus,$\int f(x) dx = \int \left(\frac{7}{5} - \frac{1}{5(5x-2)}\right) dx = \frac{7}{5}x - \frac{1}{25} \ln |5x-2| + c$.

(A) Let $I = \int \frac{x^3+2x}{x^4+4} dx$.
Split the integral into two parts: $I = \int \frac{x^3}{x^4+4} dx + \int \frac{2x}{x^4+4} dx$.
For the first part,let $u = x^4+4$,then $du = 4x^3 dx$,so $\int \frac{x^3}{x^4+4} dx = \frac{1}{4} \log|x^4+4| + C_1$.
For the second part,let $t = x^2$,then $dt = 2x dx$,so $\int \frac{2x}{(x^2)^2+4} dx = \int \frac{dt}{t^2+2^2} = \frac{1}{2} \tan^{-1}(\frac{t}{2}) + C_2 = \frac{1}{2} \tan^{-1}(\frac{x^2}{2}) + C_2$.
Combining these,$I = \frac{1}{4} \log(x^4+4) + \frac{1}{2} \tan^{-1}(\frac{x^2}{2}) + C$.
Note that $\frac{1}{4} \log(x^4+4) = \frac{1}{2} \log(\sqrt{x^4+4}) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2} \times 2) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2}) + \frac{1}{2} \log(2)$.
Absorbing the constant $\frac{1}{2} \log(2)$ into $C$,we get $I = \frac{1}{2} \left[ \tan^{-1}(\frac{x^2}{2}) + \log(\frac{\sqrt{x^4+4}}{2}) \right] + C$.

(D) We have,$f\left(\frac{t+1}{2 t+1}\right)=t+1$.
Let $\frac{t+1}{2 t+1}=x$.
Then $t+1=x(2t+1) \Rightarrow t+1=2tx+x$.
$t(1-2x)=x-1 \Rightarrow t=\frac{x-1}{1-2x}=\frac{1-x}{2x-1}$.
Substituting $t$ into the expression for $f(x)$:
$f(x) = \frac{1-x}{2x-1} + 1 = \frac{1-x+2x-1}{2x-1} = \frac{x}{2x-1}$.
Now,we calculate the integral:
$\int f(x) dx = \int \frac{x}{2x-1} dx = \frac{1}{2} \int \frac{2x}{2x-1} dx$.
$= \frac{1}{2} \int \frac{2x-1+1}{2x-1} dx = \frac{1}{2} \int \left(1 + \frac{1}{2x-1}\right) dx$.
$= \frac{1}{2} \left[ x + \frac{1}{2} \log |2x-1| \right] + c$.
$= \frac{x}{2} + \frac{1}{4} \log |2x-1| + c$.

(A) Let $I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x$.
Substitute $x=t^6$,so $d x=6 t^5 d t$.
Then $I=\int \frac{t^3}{t^3-t^2} \cdot 6 t^5 d t = 6 \int \frac{t^8}{t^2(t-1)} d t = 6 \int \frac{t^6}{t-1} d t$.
Using polynomial division,$\frac{t^6}{t-1} = t^5+t^4+t^3+t^2+t+1+\frac{1}{t-1}$.
Integrating term by term: $I = 6 \left[ \frac{t^6}{6} + \frac{t^5}{5} + \frac{t^4}{4} + \frac{t^3}{3} + \frac{t^2}{2} + t + \log|t-1| \right] + K$.
$I = t^6 + \frac{6}{5} t^5 + \frac{6}{4} t^4 + \frac{6}{3} t^3 + \frac{6}{2} t^2 + 6t + 6 \log|t-1| + K$.
Since $t = x^{1/6}$,$I = x + \frac{6}{5} x^{5/6} + \frac{3}{2} x^{2/3} + 2 x^{1/2} + 3 x^{1/3} + 6 x^{1/6} + \log(\sqrt[6]{x}-1)^6 + K$.
Comparing coefficients: $E = \frac{6}{5}$,$D = \frac{3}{2}$,$C = 2$,$B = 3$,$A = 6$.
Sum $A+B+C+D+E = 6 + 3 + 2 + \frac{3}{2} + \frac{6}{5} = 11 + 1.5 + 1.2 = 13.7 = \frac{137}{10}$.

(B) Let $I = \int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx$.
Simplify the integrand:
$\sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} = \frac{a+x + a-x}{\sqrt{(a-x)(a+x)}} = \frac{2a}{\sqrt{a^2-x^2}}$.
Now,the integral becomes:
$I = \int \frac{2a}{\sqrt{a^2-x^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$:
$I = 2a \int \frac{1}{\sqrt{a^2-x^2}} dx = 2a \sin^{-1}\left(\frac{x}{a}\right) + C$.

(C) We have the integral $I = \int \frac{e^x - 1}{e^x + 1} dx$.
To simplify the integrand,we can rewrite the numerator as $(e^x + 1) - 2$.
Thus,$I = \int \frac{(e^x + 1) - 2}{e^x + 1} dx$.
$I = \int \left( \frac{e^x + 1}{e^x + 1} - \frac{2}{e^x + 1} \right) dx = \int \left( 1 - \frac{2}{e^x + 1} \right) dx$.
Alternatively,multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{1 - e^{-x}}{1 + e^{-x}} dx$.
Let $u = 1 + e^{-x}$,then $du = -e^{-x} dx$,so $dx = -du / (u - 1)$.
Using the first method: $I = \int 1 dx - 2 \int \frac{1}{e^x + 1} dx$.
For $\int \frac{1}{e^x + 1} dx$,multiply by $e^{-x}/e^{-x}$: $\int \frac{e^{-x}}{1 + e^{-x}} dx$.
Let $v = 1 + e^{-x}$,then $dv = -e^{-x} dx$.
So,$\int \frac{1}{e^x + 1} dx = -\int \frac{1}{v} dv = -\log |1 + e^{-x}| = -\log |(e^x + 1)/e^x| = -\log (e^x + 1) + x$.
Thus,$I = x - 2(-\log (e^x + 1) + x) + c = x + 2 \log (e^x + 1) - 2x + c = 2 \log (e^x + 1) - x + c$.
Comparing with $f(x) + c$,we get $f(x) = 2 \log (e^x + 1) - x$.

(D) Given $f\left(\frac{2 x+3}{3 x+5}\right)=x+4$. Let $y = \frac{2 x+3}{3 x+5}$.
Then $y(3x+5) = 2x+3 \Rightarrow 3xy + 5y = 2x+3 \Rightarrow x(3y-2) = 3-5y \Rightarrow x = \frac{3-5y}{3y-2}$.
Substituting $x$ in $f(y) = x+4$,we get $f(y) = \frac{3-5y}{3y-2} + 4 = \frac{3-5y + 12y - 8}{3y-2} = \frac{7y-5}{3y-2}$.
Thus,$f(x) = \frac{7x-5}{3x-2}$.
Now,$\int f(x) dx = \int \frac{7x-5}{3x-2} dx = \int \frac{\frac{7}{3}(3x-2) - 5 + \frac{14}{3}}{3x-2} dx = \int \left( \frac{7}{3} - \frac{1/3}{3x-2} \right) dx$.
$= \frac{7}{3}x - \frac{1}{9} \ln |3x-2| + C$.
Comparing with $Ax + B \ln |3x-2| + C$,we get $A = \frac{7}{3}$ and $B = -\frac{1}{9}$.
Therefore,$3B - A = 3(-\frac{1}{9}) - \frac{7}{3} = -\frac{1}{3} - \frac{7}{3} = -\frac{8}{3}$.

(A) Given $I = \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx$.
Since $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$,we have $\sqrt{1-\sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$ and $\sqrt{1+\sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$.
For $\frac{5 \pi}{2} < x < \frac{7 \pi}{2}$,we have $\frac{5 \pi}{4} < \frac{x}{2} < \frac{7 \pi}{4}$.
In this interval,$\cos \frac{x}{2} < 0$ and $\sin \frac{x}{2} < 0$,and specifically $\cos \frac{x}{2} < \sin \frac{x}{2}$.
Thus,$\sqrt{1-\sin x} = \sin \frac{x}{2} - \cos \frac{x}{2}$ and $\sqrt{1+\sin x} = -(\cos \frac{x}{2} + \sin \frac{x}{2})$.
$I = \int (\sin \frac{x}{2} - \cos \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2}) dx = \int -2 \cos \frac{x}{2} dx = -4 \sin \frac{x}{2} + C$.
So $f(x) = -4 \sin \frac{x}{2}$.
Then $f'(x) = -2 \cos \frac{x}{2}$.
$f'(\frac{8 \pi}{3}) = -2 \cos \frac{4 \pi}{3} = -2 (-\frac{1}{2}) = 1$.

(D) Given $f(y) = \frac{3-8y}{3y-1}$.
We need to evaluate $\int \frac{3-8y}{3y-1} dy$.
Perform polynomial division or algebraic manipulation:
$\frac{3-8y}{3y-1} = \frac{-\frac{8}{3}(3y-1) + \frac{8}{3} + 3}{3y-1} = \frac{-\frac{8}{3}(3y-1) + \frac{17}{3}}{3y-1} = -\frac{8}{3} + \frac{17}{3(3y-1)}$.
Now,integrate with respect to $y$:
$\int (-\frac{8}{3} + \frac{17}{3(3y-1)}) dy = -\frac{8}{3}y + \frac{17}{3} \cdot \frac{1}{3} \log |3y-1| + C = -\frac{8}{3}y + \frac{17}{9} \log |3y-1| + C$.
Comparing this with $Ay + B \log |3y-1| + C$,we get $A = -\frac{8}{3}$ and $B = \frac{17}{9}$.
Then,$\frac{A-3B}{2} = \frac{-\frac{8}{3} - 3(\frac{17}{9})}{2} = \frac{-\frac{8}{3} - \frac{17}{3}}{2} = \frac{-\frac{25}{3}}{2} = -\frac{25}{6}$.
Wait,re-evaluating the original expression: $\frac{3-8y}{3y-1} = \frac{-\frac{8}{3}(3y-1) + 3 - \frac{8}{3}}{3y-1} = -\frac{8}{3} + \frac{1/3}{3y-1}$.
Integrating: $-\frac{8}{3}y + \frac{1}{9} \log |3y-1| + C$.
So $A = -\frac{8}{3}$ and $B = \frac{1}{9}$.
$\frac{A-3B}{2} = \frac{-\frac{8}{3} - 3(\frac{1}{9})}{2} = \frac{-\frac{8}{3} - \frac{1}{3}}{2} = \frac{-3}{2}$.

(C) Let $I = \int \frac{d x}{\cos (x+4) \cos (x+2)}$.
Multiply and divide by $\sin 2$:
$I = \frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d x$.
Since $2 = (x+4) - (x+2)$,we have:
$I = \frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cos (x+2)} d x$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin 2} \int \frac{\sin (x+4) \cos (x+2) - \cos (x+4) \sin (x+2)}{\cos (x+4) \cos (x+2)} d x$.
$I = \frac{1}{\sin 2} \int [\tan (x+4) - \tan (x+2)] d x$.
Integrating $\tan x$,we get $\log |\sec x|$:
$I = \frac{1}{\sin 2} [\log |\sec (x+4)| - \log |\sec (x+2)|] + C$.
$I = \frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right| + C$.

(A) Let $I = \int \frac{x+5}{x^2+4x+5} dx$.
We express the numerator as $x+5 = \lambda(2x+4) + \mu$.
Comparing the coefficients of $x$ and the constant terms on both sides:
$1 = 2\lambda \implies \lambda = \frac{1}{2}$
$5 = 4\lambda + \mu \implies 5 = 4(\frac{1}{2}) + \mu \implies 5 = 2 + \mu \implies \mu = 3$.
Thus,$I = \int \frac{\frac{1}{2}(2x+4) + 3}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} dx + 3 \int \frac{1}{(x+2)^2 + 1^2} dx$.
Using the formula $\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$ and $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{2} \log(x^2+4x+5) + 3 \tan^{-1}(x+2) + C$.
Comparing this with the given form $a \log(x^2+4x+5) + b \tan^{-1}(x+k) + C$,we get $a = \frac{1}{2}$,$b = 3$,and $k = 2$.

(C) We know that $\cos 2x = 2 \cos^2 x - 1$.
Substituting this into the integral,we get:
$\int \frac{2 \cos^2 x - 1}{\cos x} dx = \int (2 \cos x - \frac{1}{\cos x}) dx$.
This simplifies to $\int (2 \cos x - \sec x) dx$.
Integrating term by term,we get:
$2 \int \cos x dx - \int \sec x dx$.
The integral of $\cos x$ is $\sin x$ and the integral of $\sec x$ is $\log |\sec x + \tan x|$.
Thus,the final result is $2 \sin x - \log |\sec x + \tan x| + C$.

(A) Let $I = \int \frac{\sin ^8 x - \cos ^8 x}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$.
Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we can write the numerator as:
$\sin ^8 x - \cos ^8 x = (\sin ^4 x - \cos ^4 x)(\sin ^4 x + \cos ^4 x) = (\sin ^2 x - \cos ^2 x)(\sin ^2 x + \cos ^2 x)(\sin ^4 x + \cos ^4 x)$.
Since $\sin ^2 x + \cos ^2 x = 1$,the numerator becomes $(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)$.
Also,note that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x = 1 - 2 \sin ^2 x \cos ^2 x$.
Substituting this into the integral:
$I = \int \frac{(\sin ^2 x - \cos ^2 x)(1 - 2 \sin ^2 x \cos ^2 x)}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$.
$I = \int (\sin ^2 x - \cos ^2 x) \, dx = \int -(\cos ^2 x - \sin ^2 x) \, dx$.
Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get:
$I = -\int \cos 2x \, dx = -\frac{\sin 2x}{2} + C$.

(B) We know that $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.
Substituting this into the integral,we get:
$\int \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \int \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \, dx$.
Assuming $\cos \left(\frac{x}{2}\right) > 0$,we have:
$\sqrt{2} \int \cos \left(\frac{x}{2}\right) \, dx$.
Integrating with respect to $x$:
$\sqrt{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{1/2} + C = 2 \sqrt{2} \sin \left(\frac{x}{2}\right) + C$.

(C) To evaluate the integral $I = \int \frac{dx}{\sin x+\sqrt{3} \cos x}$,we multiply and divide by $2$:
$I = \int \frac{dx}{2(\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x)}$
Using the identity $\sin(x + \frac{\pi}{3}) = \sin x \cos \frac{\pi}{3} + \cos x \sin \frac{\pi}{3} = \frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x$,we get:
$I = \frac{1}{2} \int \frac{dx}{\sin(x + \frac{\pi}{3})} = \frac{1}{2} \int \operatorname{cosec}(x + \frac{\pi}{3}) dx$
Using the standard formula $\int \operatorname{cosec} \theta d\theta = \ln |\tan(\frac{\theta}{2})| + c$:
$I = \frac{1}{2} \ln |\tan(\frac{x + \frac{\pi}{3}}{2})| + c = \frac{1}{2} \ln |\tan(\frac{x}{2} + \frac{\pi}{6})| + c$.

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{4x-9x^2}}$,we first factor out $9$ from the square root:
$I = \int \frac{dx}{\sqrt{9(\frac{4}{9}x - x^2)}} = \frac{1}{3} \int \frac{dx}{\sqrt{\frac{4}{9}x - x^2}}$.
Next,we complete the square for the quadratic expression inside the square root:
$\frac{4}{9}x - x^2 = -(x^2 - \frac{4}{9}x) = -((x - \frac{2}{9})^2 - (\frac{2}{9})^2) = (\frac{2}{9})^2 - (x - \frac{2}{9})^2$.
Substituting this back into the integral:
$I = \frac{1}{3} \int \frac{dx}{\sqrt{(\frac{2}{9})^2 - (x - \frac{2}{9})^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{3} \sin^{-1}\left( \frac{x - 2/9}{2/9} \right) + C = \frac{1}{3} \sin^{-1}\left( \frac{9x - 2}{2} \right) + C$.

(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9-8x-4x^2}}$,we first complete the square for the quadratic expression in the denominator.
$9 - 8x - 4x^2 = 9 - (4x^2 + 8x) = 9 - 4(x^2 + 2x)$.
Adding and subtracting $1$ inside the bracket: $9 - 4(x^2 + 2x + 1 - 1) = 9 - 4(x+1)^2 + 4 = 13 - (2x+2)^2$.
Thus,the integral becomes $I = \int \frac{dx}{\sqrt{(\sqrt{13})^2 - (2x+2)^2}}$.
Using the standard formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,where $u = 2x+2$ and $du = 2dx$ (so $dx = \frac{du}{2}$):
$I = \frac{1}{2} \int \frac{du}{\sqrt{(\sqrt{13})^2 - u^2}} = \frac{1}{2} \sin^{-1}(\frac{u}{\sqrt{13}}) + C$.
Substituting $u = 2x+2$,we get $I = \frac{1}{2} \sin^{-1}(\frac{2x+2}{\sqrt{13}}) + C$.

(A) We have the integral $I = \int \sec^2 x \csc^2 x \, dx$.
Using the identities $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$,we get:
$I = \int \frac{1}{\cos^2 x \sin^2 x} \, dx$.
Since $1 = \sin^2 x + \cos^2 x$,we can write:
$I = \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} \, dx$.
$I = \int \left( \frac{\sin^2 x}{\cos^2 x \sin^2 x} + \frac{\cos^2 x}{\cos^2 x \sin^2 x} \right) \, dx$.
$I = \int (\sec^2 x + \csc^2 x) \, dx$.
Integrating term by term,we get:
$I = \tan x - \cot x + C$.