It's called "partial fractions". If you are expected to do this kind of problem, you should have already seen that. 1- x^4= (1- x^2)(1+ x^2)= (1- x)(1+ x)(1+ x^2) so the fraction can be written <br />
\frac{2x}{1- x^4}= \frac{A}{1- x}+ \frac{B}{1+ x}+ \frac{Dx+ C}{1+ x^2}<br />
There are a number of different wasy of determing the values of A, B, C, and D from that. For example, if you multiply both sides of the equation by 1- x^4 you get a polynomial equation that must be true for all x. Setting corresponding coefficients equal will give you four equations for A, B, D, and D. More simply, taking x= 1, -1, 0, and, say, 2 will give you four equations.<br />
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However, Bohrok's suggestion of substituting u= x<sup>2</sup> first is simpler. Once you have it reduced to <br />
\int \frac{du}{1- u^2}<br />
you can use "partial fractions" to write it as <br />
\int \frac{Adu}{1- u}+ \int \frac{Bdu}{1+ u}