Integrate $x^2 \sqrt[]{3+5x^2}dx$ (preferably) using substitution

Question

Today we went over solving integrals with tables. My task is to integrate the following:

$\int x^2 \sqrt[]{3+5x^2}dx$

In the back of the book, I am provided with over $100$ integrals. I believe this is in the form of:

$\int u \sqrt[]{a + bu} ~du = \frac{2}{15b^2}\left(3bu - 2a\right)\left(a + bu\right)^{\frac{3}{2}} + C$

So, $a = 3$ and $b = 5$.

Now, if I let $u = x$, then $du = dx$, but that does not sound right because then $a + bu$ would be $3 + 5x$.

But if I let $u = x^2$, then $du = 2xdx \rightarrow \frac{1}{2}du = xdx$. That does not sound right because there is no $x$.

Answer 1

The hint:

Let $x=\frac{1}{2}\sqrt{\frac{3}{5}}(e^t-e^{-t}).$

Thus, $$t=\ln\left(\sqrt{\frac{5}{3}}x+\sqrt{\frac{5x^2}{3}+1}\right),$$ $$dx=\frac{1}{2}\sqrt{\frac{3}{5}}(e^t+e^{-t})dt$$ and $$\int x^2\sqrt{3+5x^2}dx=\frac{9}{80\sqrt5}\int\left(e^{2t}-e^{-2t}\right)^2dt.$$ Can you end it now?

Answer 2

Hint

Making the problem more general $$I=\int x^2 \sqrt{a+b x^2}\,dx$$ where $a$ and $b$ are positive, write $$b x^2=a \sinh^2(t)\implies x=\sqrt{\frac ab} \sinh(t)\implies dx=\sqrt{\frac ab} \cosh(t)\,dt$$ Replace and use some basic properties of the hyperbolic functions.