# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int(\log x)^{2} x d x$

Solution:

Let $I=\int(\log x)^{2} x d x$

Using integration by parts,

$=(\log x)^{2} \int x d x-\int \frac{d}{d x}(\log x)^{2} \int x d x$

$=(\log x)^{2} \frac{x^{2}}{2}-\int\left(2(\log x)\left(\frac{1}{x}\right) \int x d x\right) d x$

$=\frac{x^{2}}{2}(\log x)^{2}-2 \int(\log x)\left(\frac{1}{x}\right)\left(\frac{x^{2}}{2}\right) d x$

$=\frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x$

Using integration by integration by parts in second integral,

$=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int \frac{d}{d x} \log x \int x d x\right]$

We know that, $\int \mathrm{x} \mathrm{dx}=\frac{\mathrm{x}^{2}}{2}$ and $\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}=\frac{1}{\mathrm{x}}$

$=\frac{x^{2}}{2}(\log x)^{2}-\log x \frac{x^{2}}{2}-\int \frac{1}{x} \times \frac{x^{2}}{2}$

$=\frac{x^{2}}{2}(\log x)^{2}-\log x \frac{x^{2}}{2}-\frac{1}{2} \int x d x$

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

$=\frac{x^{2}}{2}(\log x)^{2}-\log x \frac{x^{2}}{2}-\frac{x^{2}}{4}+c$

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