# Int (x/a+x^2)^3 dx = ?

May 11, 2018

$\left(\frac{1}{7}\right) {x}^{7} + \left(\frac{1}{2 a}\right) {x}^{6} + \left(\frac{3}{5 {a}^{2}}\right) {x}^{5} + \left(\frac{1}{4 {a}^{3}}\right) {x}^{4} + C$

#### Explanation:

This integral can be solved by expanding it.

Note that the following is true:

${\left(\frac{x}{a} + {x}^{2}\right)}^{3} = \left(\frac{x}{a} + {x}^{2}\right) \left(\frac{x}{a} + {x}^{2}\right) \left(\frac{x}{a} + {x}^{2}\right)$
$= \left({x}^{2} / \left({a}^{2}\right) + 2 {x}^{3} / a + {x}^{4}\right) \left(\frac{x}{a} + {x}^{2}\right)$
$= {x}^{3} / \left({a}^{3}\right) + 2 {x}^{4} / \left({a}^{2}\right) + {x}^{5} / a + {x}^{4} / \left({a}^{2}\right) + 2 {x}^{5} / a + {x}^{6}$
$= {x}^{6} + 3 {x}^{5} / a + 3 {x}^{4} / \left({a}^{2}\right) + {x}^{3} / \left({a}^{3}\right)$

Making the above substitution, our integral is now:

$\int \left({x}^{6} + \left(\frac{3}{a}\right) {x}^{5} + \left(\frac{3}{{a}^{2}}\right) {x}^{4} + \left(\frac{1}{{a}^{3}}\right) {x}^{3}\right) \mathrm{dx}$
$= \left(\frac{1}{7}\right) {x}^{7} + \left(\frac{1}{2 a}\right) {x}^{6} + \left(\frac{3}{5 {a}^{2}}\right) {x}^{5} + \left(\frac{1}{4 {a}^{3}}\right) {x}^{4} + C$

If it pleases you, you can find common denominators to make this a single term, but this form looks fine to me.