int left[frac{1-log x}{1+(log x)^{2}}right]^{ | vedclass

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

Question

(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.Substituting these into th

Vedclass · Accepted Answer

$\frac{x}{1+(\log x)^{2}}+c$

Vedclass · Answer

(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.Substituting these into the integral,we get $I = \int e^{t} \left[\frac{1-t}{1+t^{2}}\right]^{2} dt$.Expanding the square,$I = \int e^{t} \frac{1+t^{2}-2t}{(1+t^{2})^{2}} dt = \int e^{t} \left[\frac{1}{1+t^{2}} - \frac{2t}{(1+t^{2})^{2}}\right] dt$.Using the standard integral formula $\int e^{t} [f(t) + f'(t)] dt = e^{t} f(t) + c$,where $f(t) = \frac{1}{1+t^{2}}$ and $f'(t) = -\frac{2t}{(1+t^{2})^{2}}$.Thus,$I = e^{t} \left(\frac{1}{1+t^{2}}\right) + c$.Substituting back $t = \log x$,we get $I = \frac{x}{1+(\log x)^{2}} + c$.

$\int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx = $

Similar Questions

$\int_0^1 \frac{x e^x}{(2+x)^3} d x$ is equal to

$\int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx =$

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

$\int \left( \frac{x^2+1}{(x+1)^2} \right) e^x \, dx = \text{ . . . . . . }$.

Find : $\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module