int frac{(x + 1)(x + log x)^2}{x} , dx = | vedclass

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

Question

(C) Let $I = \int \frac{(x + 1)(x + \log x)^2}{x} \, dx$. We can rewrite the integrand as $\int (x + \log x)^2 \left( \frac{x + 1}{x} \right) \, dx =

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{(x + 1)(x + \log x)^2}{x} \, dx$. We can rewrite the integrand as $\int (x + \log x)^2 \left( \frac{x + 1}{x} \right) \, dx = \int (x + \log x)^2 \left( 1 + \frac{1}{x} \right) \, dx$. Let $t = x + \log x$. Then,differentiating with respect to $x$,we get $dt = (1 + \frac{1}{x}) \, dx$. Substituting these into the integral,we get $I = \int t^2 \, dt$. Integrating $t^2$ with respect to $t$,we get $I = \frac{t^3}{3} + c$. Substituting back $t = x + \log x$,we get $I = \frac{1}{3}(x + \log x)^3 + c$.

$\int \frac{(x + 1)(x + \log x)^2}{x} \, dx = $

Similar Questions

$\int \tan^3 2x \sec 2x \, dx = $

$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \, dx =$

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

If $\int \sin ^5 x \, dx = \frac{-\cos ^5 x}{5} + a \cos ^3 x + b \cos x + c$,then $a + b =$

$\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module