Solve $I_n+n I_{n-1}$, if $I_n=\int(\ln x)^n d x$

Question

Aicharya · Accepted Answer

$\begin{aligned} I_n & =\int(\log x)^n \ I_n & =(\log x)^n \int d x-\int\left(\frac{d(\log x)^n}{d x} \int d x\right) d x+k \ I_n & =x(\log x)^n-\int \frac{n(\log x)^{n-1}}{x} \times x+k \ I_n & =x(\log x)^n-n I_{n-1}+k \ \Rightarrow \quad I_n & =n I_{n-1}=x(\log x)^n+k \end{aligned}$

Solve \(I_n+n I_{n-1}\), if \(I_n=\int(\ln x)^n d x\)

Solution