# How do you integrate int 3 xln x^2 dx  using integration by parts?

Feb 12, 2017

The answer is $= 3 {x}^{2} \ln \left(| x |\right) - \frac{3}{2} {x}^{2} + C$

#### Explanation:

$\int u ' v \mathrm{dx} = u v - \int u v ' \mathrm{dx}$

Here, we have

$\int 3 x \ln \left({x}^{2}\right) \mathrm{dx} = 6 \int x \ln x \mathrm{dx}$

$v = \ln x$, $\implies$, $v ' = \frac{1}{x}$

$u ' = x$, $\implies$, $u = {x}^{2} / 2$

Therefore,

$\int 3 x \ln \left({x}^{2}\right) \mathrm{dx} = 6 \left(\frac{1}{2} {x}^{2} \ln x - \int \frac{1}{2} x \mathrm{dx}\right)$

$= 3 {x}^{2} \ln x - 3 \cdot \frac{1}{2} \cdot {x}^{2}$

$= 3 {x}^{2} \ln \left(| x |\right) - \frac{3}{2} {x}^{2} + C$