# How do you integrate x ln (x + 1) dx?

Jul 12, 2016

$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} x \left(x - 2\right) + C$

#### Explanation:

$\int \setminus x \ln \left(x + 1\right) \setminus \mathrm{dx}$

$= \int \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) \ln \left(x + 1\right) \setminus \mathrm{dx}$

and by IBP : $\int u v ' = u v - \int u ' v$

$= {x}^{2} / 2 \ln \left(x + 1\right) - \int \setminus {x}^{2} / 2 \frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right)\right) \setminus \mathrm{dx}$

$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{2} \textcolor{b l u e}{\int \setminus \frac{{x}^{2}}{x + 1}} \setminus \mathrm{dx} q \quad \triangle$

for the integration in blue, we use a simple sub

$u = x + 1 , \mathrm{du} = \mathrm{dx}$

$\int \setminus \frac{{\left(u - 1\right)}^{2}}{u} \setminus \mathrm{du}$

$\int \setminus u - 2 + \frac{1}{u} \setminus \mathrm{du}$

$= {u}^{2} / 2 - 2 u + \ln u$

$= {\left(x + 1\right)}^{2} / 2 - 2 \left(x + 1\right) + \ln \left(x + 1\right)$

so $\triangle$ becomes

$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{2} \left({\left(x + 1\right)}^{2} / 2 - 2 \left(x + 1\right) + \ln \left(x + 1\right)\right) + C$

$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{4} {\left(x + 1\right)}^{2} + \left(x + 1\right) - \frac{1}{2} \ln \left(x + 1\right) + C$

$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} + 2 x + 1 - 4 x - 4\right) + C$

$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} - 2 x - 3\right) + C$

$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} - 2 x\right) + C$

$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} x \left(x - 2\right) + C$