Integral of $\frac{x^2}{\sqrt{x^2+5}}$

Question

I need help with this integral:

$$I = \int \frac{x^2}{\sqrt{x^2+5}}\, dx$$

I substituted $x = \sqrt{5}\tan{\theta}$, and reached $$I = 5\int \frac{\sin{(\theta)}^2}{\cos{(\theta)}^3}\,d\theta$$ which I'm unable to solve. For context this is in an exercise of trig substitution, and I'm not allowed to use the secant reduction formula which WolframAlpha suggests I use.

Answer 1

Hint: (Use hyperbolic trigonometric substitution to) $$ \frac{x^2}{\sqrt{x^2+5}}=\sqrt{x^2+5}-\frac{5}{\sqrt{x^2+5}}. $$

Answer 2

Hint: Substitute $y=\sqrt{x^2+5}$. Then $dy= \frac{x}{\sqrt{x^2+5}}dx$, so the integral is $\int \sqrt{y^2-5} dy$.

Answer 3

What about

$$1+\frac x{\sqrt5}=\sinh u\implies dx=\cosh u\sqrt 5\,du\implies$$

$$\int\frac{x^2}{\sqrt{x^2+5}}==\frac1{\sqrt5}\int\frac{x^2}{\sqrt{1+\left(\frac x{\sqrt 5}\right)^2}}dx=\frac1{\sqrt5}\int\frac{(\sqrt5\,\sinh u-1)^2}{\cosh u}\sqrt5\,\cosh u\,du=$$

$$\int\left(5\sinh^2u-10\sinh u+1\right)du$$

and now it is almost an immediate integral.

Answer 4

Note that$$\frac{\sin^2(\theta)}{\cos^3(\theta)}=\frac{\sin^2(\theta)\cos(\theta)}{\bigl(1-\sin^2(\theta)\bigr)^2}.$$So, you can do $\sin(\theta)=u$ and $\cos(\theta)\,\mathrm d\theta=\mathrm du$.

Answer 5

Hint. Let $t$ such that $x^2+5=(x-t)^2$, that is $x=(t^2-5)/(2t)$. Then new integral is the integral of a rational function (very easy to calculate): $$\int \frac{x^2}{\sqrt{x^2+5}}\, dx=\int \left(\frac{25}{4t^3}+\frac{t}{4}-\frac{5}{2t}\right)\, dt.$$ Finally note that $t=x+\sqrt{x^2+5}$.

Answer 6

Hint: Rewrite $\int\frac{\sin^2{(\theta)}}{\cos^3{(\theta)}}\ d \theta$ as $\int \tan^2 \theta \sec \theta \ d \theta$. Let $u = \sec \theta$; then $du = ?$ - this will reduce the integral to something trivial.