int left(frac{x-3}{x^2+9}right)^2 , dx = | vedclass

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

Question

(C) We have $I = \int \left(\frac{x-3}{x^2+9}\right)^2 \, dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} \, dx$. Splitting the integral,we get $I = \int \fr

Vedclass · Accepted Answer

$\frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) + \frac{3}{x^2+9} + c$,where $c$ is the constant of integration.

Vedclass · Answer

(C) We have $I = \int \left(\frac{x-3}{x^2+9}ight)^2 \, dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} \, dx$. Splitting the integral,we get $I = \int \frac{x^2+9}{(x^2+9)^2} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$. $I = \int \frac{1}{x^2+9} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$. The first part is $\int \frac{1}{x^2+3^2} \, dx = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight)$. For the second part,let $u = x^2+9$,then $du = 2x \, dx$,so $3 \, du = 6x \, dx$. Thus,$\int \frac{6x}{(x^2+9)^2} \, dx = \int \frac{3}{u^2} \, du = 3 \left(-\frac{1}{u}ight) = -\frac{3}{x^2+9}$. Combining these,$I = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) - \left(-\frac{3}{x^2+9}ight) + c = \frac{1}{3} 	an^{-1}\left(\frac{x}{3}ight) + \frac{3}{x^2+9} + c$.

$\int \left(\frac{x-3}{x^2+9}\right)^2 \, dx =$

Similar Questions

$\int \frac{dx}{2+\cos x} = $ (Where $C$ is a constant of integration.)

The value of $\int_{\frac{1}{3}}^1 (x - x^3)^{\frac{1}{3}} dx$ is

Let $\tan^0 x = 1$. If $\int \left( \sum_{k=0}^7 \tan^k x \right) dx = \sum_{k=1}^7 A_k \tan^k x + C$, then $\sum_{k=1}^7 A_k$ is equal to

Let $I_n = \int \sec^n x \, dx$. If $5 I_6 - 4 I_4 = f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

$\int \frac{1}{16-7 \sin ^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module