Evaluate the following integrals:

Let $I=\int \frac{\log x}{x^{n}} d x=\int \log x \frac{1}{x^{n}} d x$

Using integration by parts,

$\int \log x \frac{1}{x^{n}} d x=\log x \int \frac{1}{x^{n}} d x-\int \frac{d}{d x} \log x \int \frac{1}{x^{n}} d x$

We know that,

$\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}$

$=\log x\left(\frac{x^{1-n}}{1-n}\right)-\int \frac{1}{x}\left(\frac{x^{1-n}}{1-n}\right) d x$

$=\log x\left(\frac{x^{1-n}}{1-n}\right)-\int\left(\frac{x^{-n}}{1-n}\right) d x$

$=\log x\left(\frac{x^{1-n}}{1-n}\right)-\left(\frac{1}{1-n}\right)\left(=\log x\left(\frac{x^{1-n}}{1-n}\right)-\right)$

$=\log x\left(\frac{x^{1-n}}{1-n}\right)-\left(\frac{x^{1-n}}{(1-n)^{2}}\right)+c$