# Evaluate integral with high power under square root

How to solve this case? If only it had been $x^2$ instead of $x^4$, I would set $x=sinht$ ...

$$\int \frac{x}{\sqrt{1+x^{4}}}dx$$

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Set $$u=x^2,\ u=\sinh t,$$ then \begin{eqnarray} \int\frac{x}{\sqrt{1+x^4}}\,dx&=&\frac12\int\frac{1}{\sqrt{1+u^2}}\,du=\frac12\int\frac{\cosh t}{\sqrt{1+\sinh^2t}}\,dt\\ &=&\frac12\int\,dt=\frac12t+c=\frac12\sinh^{-1}(x^2)+c \end{eqnarray}

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Hint 1: $x^4+1 = (x^2)^2 + 1$.

Hint 2: Hover mouse over the grey box below

Substitute $x^2 = \sinh t$. Everything simplifies nicely.

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Congrats on joining the 10k club! –  Asaf Karagila Jan 4 '13 at 11:23
@AsafKaragila: Thanks! –  Clive Newstead Jan 4 '13 at 11:26

Another approach may be to set $1+x^4=t^2x^4$ according to http://www.encyclopediaofmath.org/index.php/Differential_binomial.

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