How to solve $\int x^2\sqrt{x^2 + 4}dx$

Question

How to solve $\int x^2\sqrt{x^2 + 4}dx$?

If I will make integration by parts, I will get $\int \frac{x^4}{x^2 + 4}$, and I do not know what do with this integral.

Answer 1

Substitute $x=2\sinh u$, so you get $$ I=16\int\sinh^2u\cosh^2u\,du= 4\int\sinh^22u\,du=4\int\frac{\cosh4u-1}{2}\,du $$

Answer 2

Sub $x= 2\tan\theta, dx=2\sec^2\theta d\theta$

$\int x^2\sqrt{x^2 + 4}dx$

$=\int 4\tan^2\theta \sqrt{4\tan^2\theta+4}(2\sec^2\theta) d\theta$

$=\int (4\sec^2\theta-4) (2\sec^3\theta) d\theta$

$=8\int \sec^5\theta-\sec^3\theta d\theta$

Answer 3

Let $x = 2 \tan u$ then $dx = 2 \sec^2u~ du$. Now substitute in the integral \begin{align} \int x^2 \sqrt{x^2+4} ~dx &= \int 4 \tan^2 u \sqrt{4 \tan^2u+4} ~2\sec^2u~du \\ &= 8 \int \tan^2 u \sqrt{4\sec^2u} ~\sec^2u~du \\ &= 16 \int \tan^2 u \sec u ~\sec^2u~du \\ &= 16 \int \tan^2 u \sec^3 u ~du \\ &= 16 \int (\sec^2u-1) \sec^3 u ~du \\ &= 16 \int \sec^5u - \sec^3 u ~du \\ & = 16 \left( \frac{1}{8} (3\ln(|\tan u+\sec u|)+\sec u(2\sec^2 u+3)\tan u) - \frac{1}{2}(\ln(|\tan u+\sec u|)+\sec u\tan u)\right) + C \end{align} Now, since $\tan u = \frac{x}{2}$, then $\sec u = \frac{\sqrt{x^2+4}}{2}$. Therefore \begin{align} \int x^2 \sqrt{x^2+4} ~dx &= 16 \left( \frac{1}{8} (3\ln(|\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}|)+ \frac{\sqrt{x^2+4}}{2} (2(\frac{\sqrt{x^2+4}}{2})^2 +3)\frac{x}{2}) - \frac{1}{2}(\ln(|\frac{x}{2}+ \frac{\sqrt{x^2+4}}{2}|)+ \frac{\sqrt{x^2+4}}{2} \frac{x}{2})\right)+C\\ & = 16 \left( \frac{1}{8} (3\ln(|\frac{x+\sqrt{x^2+4}}{2}|)+ \frac{x\sqrt{x^2+4}}{2} (\frac{x^2+4}{4} +3) - \frac{1}{2}(\ln(|\frac{x+\sqrt{x^2+4}}{2}|)+ \frac{x\sqrt{x^2+4}}{4})\right)+C \end{align}

Answer 4

Write :

$$\frac{x^4}{x^2+4}=x^2-4+\frac{16}{x^2+4}$$

And now integrate both the sides.