int frac{x^4+1}{x^6+1} , dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have the integral $I = \int \frac{x^4+1}{x^6+1} \, dx$. First,rewrite the numerator as $x^4+1 = (x^4-x^2+1) + x^2$. Since $x^6+1 = (x^2)^3 + 1^

Vedclass · Accepted Answer

$	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + c$

Vedclass · Answer

(D) We have the integral $I = \int \frac{x^4+1}{x^6+1} \, dx$. First,rewrite the numerator as $x^4+1 = (x^4-x^2+1) + x^2$. Since $x^6+1 = (x^2)^3 + 1^3 = (x^2+1)(x^4-x^2+1)$,we can split the integral: $\frac{x^4+1}{x^6+1} = \frac{x^4-x^2+1}{x^6+1} + \frac{x^2}{x^6+1} = \frac{1}{x^2+1} + \frac{x^2}{(x^3)^2+1}$. Now,integrate term by term: $\int \frac{1}{x^2+1} \, dx + \int \frac{x^2}{(x^3)^2+1} \, dx$. The first integral is $	an^{-1} x$. For the second integral,let $u = x^3$,then $du = 3x^2 \, dx$,so $x^2 \, dx = \frac{1}{3} du$. Thus,$\int \frac{1}{3} \frac{du}{u^2+1} = \frac{1}{3} 	an^{-1} u = \frac{1}{3} 	an^{-1} x^3$. Combining these,we get $	an^{-1} x + \frac{1}{3} 	an^{-1} x^3 + c$.

$\int \frac{x^4+1}{x^6+1} \, dx =$

Similar Questions

Let for $f(x)=7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$,$I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7 I_1 + 12 I_2$ is equal to:

$\int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx = $

If $\int \frac{\cos 4x + 1}{\cot x - \tan x} dx = k \cos 4x + c$,then $k$ is

Integrate the function $\frac{1}{1-\tan x}$.

For $x \geq 0$,$\int \sqrt{x^2+2x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module