# What is int x^4 + 9 x^3 -8 x^2 -3 x + 5 dx?

Jan 21, 2016

$\frac{1}{5} {x}^{5} + \frac{9}{4} {x}^{4} - \frac{8}{3} {x}^{3} - \frac{3}{2} {x}^{2} + 5 x + C$

#### Explanation:

Use the rule:

$\int {x}^{n} \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1} + C$

Also, recall that multiplicative constants (like the $9$ in $9 {x}^{3}$) can be brought outside of the integral, but just stay and are multiplied in, and that various antiderivatives can be added.

Also, the antiderivative of a constant, like $5$, is simply $5 x$. This is true since $5 = 5 {x}^{0}$, to which the original rule can be applied.

The antiderivative of the given function is:

${x}^{4 + 1} / \left(4 + 1\right) + \frac{9 {x}^{3 + 1}}{3 + 1} - \frac{8 {x}^{2 + 1}}{2 + 1} - \frac{3 {x}^{1 + 1}}{1 + 1} + \frac{5 {x}^{0 + 1}}{0 + 1} + C$

$= \frac{1}{5} {x}^{5} + \frac{9}{4} {x}^{4} - \frac{8}{3} {x}^{3} - \frac{3}{2} {x}^{2} + 5 x + C$