The value of I=int frac{x^2}{(a+b x)^2} ,d x | vedclass

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

Question

(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.Substituting these into the integral:$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \fr

Vedclass · Accepted Answer

$\frac{1}{b^3}\left[a+b x-2 a \log |a+b x|-\frac{a^2}{a+b x}\right]+c$,(where $c$ is the constant of integration)

Vedclass · Answer

(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.Substituting these into the integral:$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \frac{dt}{b} = \frac{1}{b^3} \int \frac{t^2-2at+a^2}{t^2} dt$$I = \frac{1}{b^3} \int (1 - \frac{2a}{t} + \frac{a^2}{t^2}) dt$$I = \frac{1}{b^3} [t - 2a \log |t| - \frac{a^2}{t}] + c$Substituting $t=a+bx$ back:$I = \frac{1}{b^3} [a+bx - 2a \log |a+bx| - \frac{a^2}{a+bx}] + c$

The value of $I=\int \frac{x^2}{(a+b x)^2} \,d x$ is

Similar Questions

$\int \sin^{\frac{-1}{2}}x \cos^{\frac{-7}{2}}x \, dx = $

Integrate the following function with respect to $x:$
$2 x \sin \left(x^{2}+1\right)$

$\int \frac{1}{x\sqrt{1 + \log x}} \, dx = $

Integrate the function: $\frac{3x^{2}}{x^{6}+1}$

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$ is equal to-

Vedclass Test Series

Exam Paper Generator

Online Exam Module