int frac{dx}{sqrt{9-8x-4x^2}} = . . . . . . | vedclass

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

Question

(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 -

Vedclass · Accepted Answer

$\frac{1}{2} \sin^{-1} (\frac{2x+2}{\sqrt{13}})$

Vedclass · Answer

(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 $I = \frac{1}{2} \sin^{-1}(\frac{2x+2}{\sqrt{13}}) + C$.

$\int \frac{dx}{\sqrt{9-8x-4x^2}} = $ . . . . . . + $C$

Similar Questions

$\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=$

Find the following integral: $\int(ax^{2} + bx + c) dx$

Find the following integral: $\int(2x - 3 \cos x + e^x) \, dx$

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: $\int \sqrt{x-3} \left(\sin^{-1}(\log x) + \cos^{-1}(\log x)\right) dx = \frac{\pi}{3}(x-3)^{3/2} + c$
Reason $(R)$: $\sin^{-1}(f(x)) + \cos^{-1}(f(x)) = \frac{\pi}{2}$ for $|f(x)| \le 1$
Choose the correct option:

If $g\left(\frac{t+1}{2 t+1}\right)=t+1$,then $\int g(x) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module