How do you integrate int (x+1)^2ln3x by integration by parts method?

Nov 3, 2016

${\left(x + 1\right)}^{3} / 3 \ln 3 x - \frac{1}{3} \left({x}^{3} / 3 + \frac{3}{2} {x}^{2} + 3 x + \ln \left\mid x \right\mid\right) + c$

Explanation:

${f}^{'} \left(x\right) = {\left(x + 1\right)}^{2} \implies f \left(x\right) = {\left(x + 1\right)}^{3} / 3$
$g \left(x\right) = \ln 3 x \implies g ' \left(x\right) = \frac{1}{x}$

$\int {\left(x + 1\right)}^{2} \ln 3 x \setminus \mathrm{dx} = {\left(x + 1\right)}^{3} / 3 \ln 3 x - \frac{1}{3} \int {\left(x + 1\right)}^{3} \setminus \frac{1}{x} \setminus \mathrm{dx} =$
$= {\left(x + 1\right)}^{3} / 3 \ln 3 x - \frac{1}{3} \int \frac{{x}^{3} + 3 {x}^{2} + 3 x + 1}{x} \setminus \mathrm{dx} =$
$= {\left(x + 1\right)}^{3} / 3 \ln 3 x - \frac{1}{3} \int {x}^{2} + 3 x + 3 + \frac{1}{x} \setminus \mathrm{dx} =$
$= {\left(x + 1\right)}^{3} / 3 \ln 3 x - \frac{1}{3} \left({x}^{3} / 3 + \frac{3}{2} {x}^{2} + 3 x + \ln \left\mid x \right\mid\right) + c$