Question

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

Answer 1

Let $x+\log x=t$

Differentiating both sides w.r.t $x$

$1+\frac{1}{x}=\frac{dt}{dx}$

$\frac{x+1}{x}=\frac{dt}{dx}$

$dx=\left(\frac{x}{x+1}\right)dt$

Integrating the function

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

Put the value of $x+\log x=t$ and $dx=\left(\frac{x}{x+1}\right)dt$

$=\int \frac{(x+1)t^2}{x}\frac{x}{x+1}dt$

$=\int t^2.dt$

$=\frac{t^{2+1}}{2+1}+c$

$=\frac{t^3}{3}+c$

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