Question

If $\int \frac{x^x(1+\log x)dx}{x^{x+1}}$ $=\frac{1}{2}{(1+\log x)}^n+c$, then find the value of $n$.

Answer 1

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

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

Let $1+\log x=t$

differentiating w.r.t t, we have

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

$\Rightarrow \int tdt=\frac{t^2}{2}+c$

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

Hence, the answer is $\frac{1}{2}{(1+\log x)}^2+c.$