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We have to solve the integration

$\int \frac {dx}{(x^4 -1)^2}$

In this tried to substitute x = tan $\theta$

But by doing this I am not able to proceed .

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    You can use partial fractions and then integrate the sum term by term. – 2017-01-03
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    @MarvinF. In this I think partial fractions will become too long – 2017-01-03
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    Sure, it can be tedious to do the partial fractions, according to Wolframalpha $$\frac{1}{(x^4-1)^2}=\frac{1}{4(x^2+1)}+\frac{1}{4(x^2+1)^2}+\frac{3}{16(x+1)}+\frac{1}{16(x+1)^2}-\frac{3}{16(x-1)}+\frac{1}{16(x-1)^2}$$ but after this integrating becomes quite simple. – 2017-01-03

2 Answers 2

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HINT:

Write the numerator as $$4=(x^2+1+x^2-1)^2$$

Then again $2=x^2+1+x^2-1$

$$\dfrac4{(x^2-1)^2(x^2+1)^2}=\dfrac{(x^2+1+x^2-1)^2}{(x^2-1)^2(x^2+1)^2}$$

$$=\dfrac{(x^2+1)^2+(x^2-1)^2+2(x^2+1)(x^2-1)}{(x^2-1)^2(x^2+1)^2}$$

$$=\dfrac1{(x^2-1)^2}+\dfrac1{(x^2+1)^2}+\dfrac2{(x^2+1)(x^2-1)}$$

Now $\dfrac2{(x^2+1)(x^2-1)}=\dfrac{(x^2+1)-(x^2-1)}{(x^2+1)(x^2-1)}$

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    Could not understand what you have written – 2017-01-03
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    @koolman, Can you take it from here? – 2017-01-03
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try to split the denominator $$(x^4 - 1)^2 =((x^2 - 1)*(x^2 + 1))^2$$ You can then write the integral as $$\int \frac {dx}{(x^4 -1)^2} = \frac{1}{4} \int [{\frac {1}{(x^2 -1)} - {\frac {1}{(x^2 +1)}}}]^2dx$$ open up the square $$\int \frac {dx}{(x^4 -1)^2} = \frac{1}{4} \int {\frac {1}{(x^2 -1)^2} + {\frac {2}{(x^2 -1)*(x^2 +1)} + {\frac {1}{(x^2 +1)^2}}}}dx$$

For the last term use $x = tan\theta$ and use the identity $2cos^2\theta = 1 + cos2\theta$. For the middle term split into partial fractions and integrate (one part by splitting up again into partial fractions and the other by substituting by $tan\theta$)

The first term follows the pattern of the current integral using power 2 instead of 4. So follow the same steps used.