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

Jul 2, 2015

Using the following identity:
${\tan}^{2} x + 1 = {\sec}^{2} x$

the following substitution works out:
$x = 2 \tan \theta$
$\mathrm{dx} = 2 {\sec}^{2} \theta d \theta$
${x}^{3} = 8 {\tan}^{3} \theta$
$\sqrt{{x}^{2} + 4} = 2 \sec \theta$

Now we get:

$\int 8 {\tan}^{3} \theta \cdot 2 \sec \theta \cdot 2 {\sec}^{2} \theta d \theta$

$= 32 \int {\tan}^{3} \theta {\sec}^{3} \theta d \theta$

$= 32 \int {\tan}^{2} \theta \tan \theta {\sec}^{3} \theta d \theta$

Now let's separate it so hopefully we get something we can do with u-substitution.
$= 32 \int \left({\sec}^{2} \theta - 1\right) \tan \theta {\sec}^{3} \theta d \theta$

$= 32 \int {\sec}^{5} \theta \tan \theta - {\sec}^{3} \theta \tan \theta d \theta$

Notice how we can force this to look like ${u}^{4} \mathrm{du} - {u}^{2} \mathrm{du}$:
$= 32 \int {\left(\sec \theta\right)}^{4} \sec \theta \tan \theta - {\left(\sec \theta\right)}^{2} \sec \theta \tan \theta d \theta$

Now, we can do another substitution:
$u = \sec \theta$
$\mathrm{du} = \sec \theta \tan \theta d \theta$

This gives:
$= 32 \int {u}^{4} - {u}^{2} \mathrm{du}$

$= 32 \left({u}^{5} / 5 - {u}^{3} / 3\right)$

$= 32 \left({\sec}^{5} \frac{\theta}{5} - {\sec}^{3} \frac{\theta}{3}\right)$

Now, since $2 \sec \theta = \sqrt{{x}^{2} + 4} \to \sec \theta = \frac{\sqrt{{x}^{2} + 4}}{2}$:

$= 32 \left({\left({x}^{2} + 4\right)}^{\frac{5}{2}} / \left({2}^{5} \cdot 5\right) - {\left({x}^{2} + 4\right)}^{\frac{3}{2}} / \left({2}^{3} \cdot 3\right)\right)$

For convenience I've purposefully multiplied the second fraction by $\frac{4}{4}$.
$= \cancel{32} \left({\left({x}^{2} + 4\right)}^{\frac{5}{2}} / \left(\cancel{32} \cdot 5\right) - \frac{4 {\left({x}^{2} + 4\right)}^{\frac{3}{2}}}{\cancel{\left(4 \cdot 8\right)} \cdot 3}\right)$

$= \textcolor{g r e e n}{{\left({x}^{2} + 4\right)}^{\frac{5}{2}} / \left(5\right) - \frac{4 {\left({x}^{2} + 4\right)}^{\frac{3}{2}}}{3} + C}$

If you want, you can stop here, but Wolfram Alpha gives you a different answer, so I'll show you how to get there too if you use that to check. Alternatively you could say "Is A = B" and it'll check for you, where A = one answer and B = the one you're checking.

Factor out $\frac{1}{5}$:
$= \frac{1}{5} \left[{\left({x}^{2} + 4\right)}^{\frac{5}{2}} - \frac{20 {\left({x}^{2} + 4\right)}^{\frac{3}{2}}}{3}\right]$

Factor out $\frac{1}{3}$:
$= \frac{1}{15} \left[3 {\left({x}^{2} + 4\right)}^{\frac{5}{2}} - 20 {\left({x}^{2} + 4\right)}^{\frac{3}{2}}\right]$

Factor out ${\left({x}^{2} + 4\right)}^{\frac{3}{2}}$:
$= \frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \left[3 \left({x}^{2} + 4\right) - 20\right]$

$= \frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \left[3 {x}^{2} + 12 - 20\right]$
$= \textcolor{b l u e}{\frac{1}{15} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} \left[3 {x}^{2} - 8\right] + C}$