How to evaluate the following integral without using partial fraction?

Question

$$\int \frac{dx}{(x^4-1)^2} $$ One way would be applying partial fractions. But Its very tedious. Can you suggest any other way ?

Answer 1

$$\int\dfrac{dx}{(x^4-1)^2}=\dfrac14\int\dfrac1{x^3}\dfrac{4x^3}{(x^4-1)^2}dx$$

$$=\dfrac1{4x^3}\int\dfrac{4x^3}{(x^4-1)^2}dx-\dfrac14\int\left(\dfrac{d(x^{-3})}{dx}\cdot\int\dfrac{4x^3}{(x^4-1)^2}dx\right)dx$$

$$=\dfrac1{4x^3}\left(-\dfrac1{x^4-1}\right)-\dfrac34\int\dfrac1{x^4(x^4-1)}dx$$

Now $$\int\dfrac1{x^4(x^4-1)}dx=\int\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}dx=?$$

and $$\int\dfrac2{x^4-1}dx=\int\dfrac{x^2+1-(x^2-1)}{(x^2-1)(x^2+1)}dx=??$$

Can you take it from here?

Answer 2

Let $$I=\int\frac{dx}{x^4-1}$$ and $$J=\int\frac{dx}{(x^4-1)^2}$$

Use parts for $I$ we get

$$I=\frac{x}{x^4-1}+4\int\frac{x^4dx}{(x^4-1)^2}$$ $\implies$

$$I=\frac{x}{x^4-1}+4\int\frac{x^4-1+1dx}{(x^4-1)^2}$$ $\implies$

$$I=\frac{x}{x^4-1}+4I+4J$$

Originally $I$ can be integrated easily by partial fractions.

Answer 3

Let $$\displaystyle I = \int\frac{1}{(x^4-1)^2}dx = \frac{1}{4}\int \left[\frac{-3}{(x^4-1)}+\frac{3x^4+1}{(x^4-1)^2}\right]dx$$

$$\displaystyle I = \frac{1}{4}\int\left[\frac{-3}{2(x^2-1)}+\frac{3}{2(x^2+1)}+\frac{3x^2+\frac{1}{x^2}}{\left(x^3-\frac{1}{x}\right)^2}\right]dx$$

Now Here $\bf{1^{st}}$ and $\bf{2^{nd}}$ can be Calculated Direct Formulas and for third put

$$\displaystyle \left(x^3-\frac{1}{x}\right) = t\;,$$ Then $$\displaystyle \left(3x^2+\frac{1}{x^2}\right)dx = dt$$

so we get $$\displaystyle I = \frac{3}{16}\ln \left|\frac{1+x}{1-x}\right|+\frac{3}{8}\tan^{-1}(x)-\frac{1}{4}\cdot \frac{x}{(x^4-1)}+\mathcal{C}$$