# How do you find the antiderivative of int x^3/sqrt(4x^2-1)dx?

Nov 5, 2016

$\frac{\left(2 {x}^{2} + 1\right) \sqrt{4 {x}^{2} - 1}}{24} + C$

#### Explanation:

$I = \int {x}^{3} / \sqrt{4 {x}^{2} - 1} \mathrm{dx}$

Let $u = 4 {x}^{2} - 1$ so that $\mathrm{du} = 8 x \mathrm{dx}$. Also note that ${x}^{2} = \frac{u + 1}{4}$, which will be useful in a second:

$I = \frac{1}{8} \int \frac{{x}^{2} \left(8 x\right)}{\sqrt{4 {x}^{2} - 1}} \mathrm{dx}$

Substituting in our $8 x \mathrm{dx}$ and ${x}^{2}$ terms we have:

$I = \frac{1}{8} \int \frac{\frac{u + 1}{4}}{\sqrt{u}} \mathrm{du} = \frac{1}{32} \int \frac{u + 1}{\sqrt{u}} \mathrm{du}$

$I = \frac{1}{32} \int \left({u}^{\frac{1}{2}} + {u}^{- \frac{1}{2}}\right) \mathrm{du}$

Now integrating using the power rule for integration:

$I = \frac{1}{32} \left({u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right) = \frac{1}{32} \left(\frac{2}{3} {u}^{\frac{3}{2}} + 2 {u}^{\frac{1}{2}}\right)$

$I = \frac{1}{48} {u}^{\frac{3}{2}} + \frac{1}{16} {u}^{\frac{1}{2}} = \frac{{u}^{\frac{3}{2}} + 3 {u}^{\frac{1}{2}}}{48} = \frac{{u}^{\frac{1}{2}} \left(u + 3\right)}{48}$

Now since $u = 4 {x}^{2} - 1$:

$I = \frac{\sqrt{4 {x}^{2} - 1} \left(4 {x}^{2} - 1 + 3\right)}{48} = \frac{\sqrt{4 {x}^{2} - 1} \left(4 {x}^{2} + 2\right)}{48}$

So:

$I = \frac{\sqrt{4 {x}^{2} - 1} \left(2 {x}^{2} + 1\right)}{24} + C$