# How do you find the indefinite integral of int x^3(x^4+3)^2dx?

Sep 28, 2015

${\left({x}^{4} + 3\right)}^{2} / 12 + C$

#### Explanation:

Let ${x}^{4} + 3 = u$, so that ${x}^{3} \mathrm{dx} = \frac{1}{4} \mathrm{du}$ On substituting, the given integral becomes $\int \frac{1}{4} {u}^{2} \mathrm{du}$

=${u}^{3} / 12 + c$

=${\left({x}^{4} + 3\right)}^{2} / 12 + C$

Sep 28, 2015

The answer is ${x}^{12} / 12 + 3 {x}^{8} / 4 + 9 {x}^{4} / 4 + c$.

#### Explanation:

First of all, expand the square: using the formula

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

we have that

${\left({x}^{4} + 3\right)}^{2} = {x}^{8} + 6 {x}^{4} + 9$

Now we can multiply the expanded square for ${x}^{3}$, obtaining

${x}^{3} \cdot \left({x}^{8} + 6 {x}^{4} + 9\right) = {x}^{11} + 6 {x}^{7} + 9 {x}^{3}$

Now this quantity is easy to integrate, because of the linearity of the integral, which means that the integral of the sum is the sum of the integrals:

$\setminus \int {x}^{11} + 6 {x}^{9} + 9 {x}^{3} \mathrm{dx} = \setminus \int {x}^{11} \mathrm{dx} + \setminus \int 6 {x}^{7} + \setminus \int 9 {x}^{3} \mathrm{dx}$

and each of these integrals can be done using the same rule, i.e.

$\setminus \int a {x}^{n} \mathrm{dx} = \frac{a {x}^{n + 1}}{n + 1}$.

So: $\setminus \int {x}^{11} = {x}^{12} / 12$,
$\setminus \int 6 {x}^{7} = 6 {x}^{8} / 8 = 3 {x}^{8} / 4$, and
$\setminus \int 9 {x}^{3} = 9 {x}^{4} / 4$.

The answer is thus ${x}^{12} / 12 + 3 {x}^{8} / 4 + 9 {x}^{4} / 4 + c$.