Evaluate the integral using a specified substitution

Question

I'm asked to evaluate the following integral: $$\int \frac{x^2}{\sqrt{x^2+4}}dx$$ using

a) Trigonometric substitution

b) The substitution $x=t-t^{-1}$.

I did part a) just fine. The Integral is $\frac{1}{2}x\sqrt{x^2+4}-2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C$ which is correct.

However, I can't figure out how to do part b). If I let $x=t-t^{-1}$, then I can make the following substitutions:

$x=t-t^{-1} \rightarrow x=t-\frac{1}{t} \rightarrow x=\frac{t^2-1}{t}$ or, $x^2=\frac{(t^2-1)^2}{t^2}$.

Differentiating, this gives $dx=(1+t^{-2})dt \rightarrow dx=\left( 1+\frac{1}{t^2} \right) dt \rightarrow dx = \left(\frac{t^2+1}{t^2}\right) dt$

So now, my integral becomes:

$$\int \frac{x^2}{\sqrt{x^2+4}}dx \rightarrow \int \frac{(t^2-1)^2}{t^2\sqrt{\frac{(t^2-1)^2}{t^2}+4}} \frac{t^2+1}{t^2} dt$$

However, I'm not really sure how to proceed from here... Can someone lend a hand and perhaps guide me through this? Thanks!

I should theoretically get the same answer as part a.

Answer 1

Note that\begin{align}\sqrt{\left(t-\frac1t\right)^2+4}&=\sqrt{t^2+2+\frac1{t^2}}\\&=t+\frac1t.\end{align}Can you take it from here?

Answer 2

\begin{align} \int \frac{x^2}{\sqrt{x^2+4}}dx &= \int \frac{(t^2-1)^2}{t^2\sqrt{\frac{(t^2-1)^2}{t^2}+4}} \frac{t^2+1}{t^2} dt\\ &= \int \frac{(t^2-1)^2 }{t^2\sqrt{\frac{(t^2-1)^2+ 4t^2}{t^2}}} \frac{t^2+1}{t^2} dt\\ &= \int \frac{(t^2-1)^2 }{t^2\frac{1}{t}\sqrt{(t^2-1)^2+ 4t^2}} \frac{t^2+1}{t^2} dt\\ &= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^2-1)^2+ 4t^2}} (t^2+1) dt\\ &= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^4-2t^2 + 1)+ 4t^2}} (t^2+1) dt\\ &= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^4+2t^2 + 1)}} (t^2+1) dt\\ &= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^2 + 1)^2}} (t^2+1) dt\\ \end{align} ...and you can probably take it from here.