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

Question

I'm trying to find $\displaystyle\int\frac{1}{(x^2-1)^2}dx$. Partial fractions just gives me the same fraction. When I try trig substitution, I end up with $\dfrac{\cos^2(x)}{\sin^3(x)}$. What am I doing wrong? What can I do?

Answer 1

$$\dfrac4{(x^2-1)^2}=\dfrac{\{(x+1)-(x-1)\}^2}{(x-1)^2\cdot(x+1)^2}=\dfrac1{(x-1)^2}+\dfrac1{(x+1)^2}-\dfrac2{(x+1)(x-1)}$$

Use $2=(x+1)-(x-1)$ for the last case

Answer 2

Integrating by part,

$$\int\dfrac1x\cdot\dfrac x{(x^2-1)^2}dx=\dfrac1x\cdot\int\dfrac x{(x^2-1)^2}dx-\int\left(\dfrac{d(1/x)}{dx}\cdot\int\dfrac x{(x^2-1)^2}dx\right)dx$$

Now $\displaystyle\int\dfrac x{(x^2-1)^2}dx=?$

Can you take it from here?

Answer 3

Then you have done the partial fractions wrong. Partial fractions can't "give the same thing" here. Since $x^2- 1= (x- 1)(x+ 1)$, $(x^2- 1)^2= (x- 1)^2(x+ 1)^2$ so the partial fractions form is $\frac{1}{(x^2- 1)^2}= \frac{A}{x- 1}+ \frac{B}{(x- 1)^2}+ \frac{C}{x+ 1}+ \frac{D}{(x+ 1)^2}$.

Answer 4

Hint:

By educated guess, let us try

$$F(x)=\frac x{x^2-1}.$$

Taking the derivative, we find

$$F'(x)=\frac{x^2-1-2x^2}{(x^2-1)^2}=-\frac1{x^2-1}-\frac2{(x^2-1)^2}.$$

The antiderivative of $\dfrac1{1-x^2}$ is known to be the inverse hyperbolic tangent.

Answer 5

Let $x=\sec t$ and $dx=\sec t\tan t dt$

$$\int\frac{1}{(x^2-1)^2}dx=\int\frac{\sec t\tan t}{(\sec^2t-1)^2}dt=\int\frac{\sec t\tan t}{\tan^4t}dt=\int\frac{\sec t}{\tan^3t}dt=\int\frac{\cos^2t}{\sin^3t}dt=\int\frac{1-\sin^2t}{\sin^3t}dt=\int\frac{1}{\sin^3t}dt-\int\frac{1}{\sin t}dt$$

You can use trigonometric identities and commonly known integrals for the final steps.

Answer 6

We have: $$ \frac{1}{1+x}+\frac{1}{1-x} = \frac{2}{1-x^2} \tag{1}$$ so, by squaring both sides: $$ \frac{1}{(1+x)^2}+\frac{1}{(1-x)^2}+\frac{1}{1+x}+\frac{1}{1-x} = \frac{4}{(1-x^2)^2}.\tag{2} $$ It follows that: $$ \int \frac{dx}{(1-x^2)^2} = C+\frac{1}{4}\left(-\frac{1}{1+x}+\frac{1}{1-x}+\log(1+x)-\log(1-x)\right).\tag{3}$$