Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|cite|improve this question

3 Answers 3

up vote 1 down vote accepted

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}

share|cite|improve this answer

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.

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.