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# Evaluate the following $\int\frac{x}{x^4-1}dx$

Can you answer this question?

Toolbox:
• $\large\int \frac{dx}{x^2-a^2}= \log |x+\sqrt {x^2+a^2}|+c$
Let $\large\int\frac{x}{x^4-1}dx$
Let $x^2=t$
differentiating with respect to t we get,
$2xdx=dt$
$=> xdx=\large\frac{dt}{2}$
Now substituting for $x^2$ and $xdx$ we get
Therefore $I=\large\frac{dt/2}{t^2-1}=\frac{1}{2} \int \frac{dt}{t^2-1}$
This is of the form $\large\frac{dx}{x^2-a^2}=\log |x+\sqrt {x^2-a^2}|+c$
Hence on integraing we get,
$\large\frac{1}{2} \int \frac{dt}{t^2-1}=\log |t+\sqrt {t^2-1}|+c$
Substituting for t we get
$\log |x^2+\sqrt {x^4-1}|+c$
answered Apr 16, 2013 by