# How do you evaluate the integral int ln(x^2+x^3)?

Jan 14, 2017

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

#### Explanation:

We have the integral:

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

We should apply integration by parts, which takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Where $\int \ln \left({x}^{2} + {x}^{3}\right) \mathrm{dx} = \int u \mathrm{dv}$, let $u = \ln \left({x}^{2} + {x}^{3}\right)$ and $\mathrm{dv} = \mathrm{dx}$.

Differentiating $u$ and integrating $\mathrm{dv}$ gives:

$\left\{\begin{matrix}u = \ln \left({x}^{2} + {x}^{3}\right) \text{ "=>" "du=(2x+3x^2)/(x^2+x^3)dx=(2+3x)/(x+x^2)dx \\ dv=dx" "=>" } v = x\end{matrix}\right.$

So:

$I = u v - \int v \mathrm{du}$

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

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

Either perform long division on $\frac{3 x + 2}{x + 1}$ or do the following rewriting: $\frac{3 x + 2}{x + 1} = \frac{3 \left(x + 1\right) - 1}{x + 1} = 3 - \frac{1}{x - 1}$. So:

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

$I = x \ln \left({x}^{2} + {x}^{3}\right) - \left(3 x - \ln \left(\left\mid x - 1 \right\mid\right)\right)$

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