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?
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Not so much of a difference, but $t^2+2t-1 = (t+1)^2-2$, so you could substitute $t+1 = \sqrt{2}\cosh u$, which may be a little nicer. – Daniel Fischer Mar 22 '14 at 14:23
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I would not be surprised if it is not doable by elementary functions. – André Nicolas Mar 22 '14 at 15:59
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$$I=\int\sqrt{x^4+2x^2-1}\,x\,dx$$
$$I=\int\sqrt{(x^2+1)^2-2}\,x\,dx$$
Let $u=x^2+1\implies x\,dx=\frac{du}{2}$
$$\frac12\int\sqrt{u^2-2}du$$
Doable now?
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.
Guy
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