Integrate the function frac{(1+log x)^{2}}{x} | vedclass

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

Question

(A) Let $1+\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\i

Vedclass · Accepted Answer

$\frac{(1+\log x)^{3}}{3}+C$

Vedclass · Answer

(A) Let $1+\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\int \frac{(1+\log x)^{2}}{x} dx = \int t^{2} dt$. Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$,we get: $\int t^{2} dt = \frac{t^{3}}{3} + C$. Substituting $t = 1+\log x$ back into the expression,we get: $\frac{(1+\log x)^{3}}{3} + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{(1+\log x)^{2}}{x}$.

Similar Questions

If $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$,then $e^{f(x)}=$

$\int \frac{dx}{x(x^7 + 1)} = $

Evaluate the integral $\int \frac{d x}{(x+100) \sqrt{x+99}} = f(x) + c$. Find $f(x)$.

$\int \sin \sqrt{x} \, dx = $

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module