# How do you find the integral of (x^3 + 1) lnxdx?

Apr 30, 2018

$\left({x}^{4} / 4 + x\right) \left(I n x\right) - {x}^{4} / 16 + x + C$

#### Explanation:

Let u= $I n x$
And dv=${x}^{3} + 1$ SO v=$\left({x}^{4} / 4 + x\right)$

Integration by parts is given by = $v u - \int v \mathrm{du}$

$\int \left({x}^{3} + 1\right) I n x \mathrm{dx}$ = $\left({x}^{4} / 4 + x\right) \left(I n x\right) - \int \left(x\right) \left({x}^{3} / 4 + 1\right) \times \frac{1}{x} \mathrm{dx}$

=$\left({x}^{4} / 4 + x\right) \left(I n x\right) - \int \left({x}^{3} / 4 + 1\right) \mathrm{dx}$

=$\left({x}^{4} / 4 + x\right) \left(I n x\right) - {x}^{4} / 16 + x + C$