Integrate irrational function

Question

I need to solve an indefinite integral $$ I=\int\sqrt{x^4+2x^2-1}\,x\,dx. $$ Substituting $x^2=t$ yields $$ I=\frac{1}{2}\int\sqrt{t^2+2t-1}\,dt=\frac{1}{2}\int\sqrt{t+1-\sqrt{2}}\sqrt{t+1+\sqrt{2}}\,dt. $$ Now substituting $s=\sqrt{t+1-\sqrt{2}}\,$ yields $$ I=\int\sqrt{s^2+2\sqrt{2}}\,s^2ds $$ which opens way to solving the integral in hyperbolic functions, but it seems to me too much complicated. My question is: Is this the only way? Maybe there is something else, not so complicated?

Answer 1

$$I=\int\sqrt{x^4+2x^2-1}\,x\,dx$$

$$I=\int\sqrt{(x^2+1)^2-2}\,x\,dx$$

Let $u=x^2+1\implies x\,dx=\frac{du}{2}$

$$\frac12\int\sqrt{u^2-2}du$$

Doable now?

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In case you are still stuck, proceed like so.

$u=\sqrt{2}\sec(\theta)\implies du=\sqrt{2}\sec(\theta)\tan(\theta)d\theta$

$$\frac12\int \sqrt{2}\sec(\theta)\tan^2(\theta)\,d\theta$$

EDIT:Made a bad mistake here. Proceed through integration by parts.