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

Question

How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?

Note that this is a Q&A post and I've presented my solution below.

Let $$x=\sqrt{\sec({2\tan^{-1}}t)}$$

$$\Rightarrow dx=\frac{2\tan^{2}(2\tan^{-1}t)}{2(1+t^2)\sqrt{\sec(2\tan^{-1}t)}}$$

$$\Rightarrow xdx=\frac{\tan^2(2\tan^{-1}t)}{1+t^2}$$

Substituting in the original integral- $$\int \frac{2\tan^2(2\tan^{-1}t)}{(1+t^2)(1-\sec(2\tan^{-1}t))(\sqrt{\sec^2(2\tan^{-1}t)-1})}dt$$ $$= \int \frac{2\tan(2\tan^{-1}t)}{(1+t^2)(1-\sec(2\tan^{-1}t))}dt$$ $$= \int \frac{4t(1-t^2)}{(-2t^2)(1-t^4)}dt$$ $$= -2\int \frac{dt}{t(1+t^2)}$$ $$= -2\left(\log t-\frac{1}{2}\log(t^2+1)\right)+C$$

Answer 1

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

$\Rightarrow\int \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx$

Now, one substitution that works but is rather tricky to find is,

Let $u=\sqrt{\frac{x^2+1}{x^2-1}}$

$\Rightarrow \frac{1}{2}(\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{(x^2-1)2x-(x^2+1)2x}{(x^2-1)^2}dx=du$

$\Rightarrow (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{-2x}{(x^2-1)^2}dx=du$

$\Rightarrow \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx=du$

The integral reduces to

$\int du=u+c=\sqrt{\frac{x^2+1}{x^2-1}}+c$

Thus $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx=\sqrt{\frac{x^2+1}{x^2-1}}+c$

Answer 2

Letting $x^2=\sec \theta$ transform the integral into \begin{aligned} I & =\int \frac{\sec \theta \tan \theta d \theta}{(1-\sec \theta) \tan \theta} \\ & =\int \frac{1}{\cos \theta-1} d \theta \\ & =\int \frac{\cos \theta+1}{-\sin ^2 \theta} d \theta \\ & =-\int \cot \theta \csc \theta d \theta-\int \csc ^2 \theta d \theta \\ & =\csc \theta+\cot \theta+C \\ & =\frac{x^2+1}{\sqrt{x^4-1}}+C \end{aligned}

Answer 3

Substitute $x^2=t$:

$$\int \frac{\mathrm dt}{(1-t)\sqrt{t^2-1}}$$

Now, put $1-t=\frac1{u}$. Note that $u=\frac1{1-x^2}<0$ as there is $\sqrt{x^4-1}=\sqrt{(x^2-1)(x^2+1)}$ in the denominator of the integrand, so $x^2>1$:

$$\int \frac{\mathrm du}{u\sqrt{(1-\frac1{u})^2-1}}$$ Now since $u<0$, we can take $-u$ inside the square root and not $u$: $$=-\int \frac{\mathrm du}{\sqrt{(u-1)^2-u^2}}$$ $$=-\int \frac{\mathrm du}{\sqrt{1-2u}}$$ $$=\sqrt{1-2u}+C$$ $$=\sqrt{1+\frac{2}{x^2-1}}+C$$ $$=\sqrt{\frac{x^2+1}{x^2-1}}+C$$

Answer 4

Substitute $x^2=\cosh 2t$ to integrate \begin{align} \int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx =-\int \text{csch}^2 t \ dt =\coth t +C \end{align}

Answer 5

Another way to evaluate the OP’s integral is using Euler’s substitutions.

By letting $\;t=x^2\,,\;$ we get that

$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt\;.$

Now, we will apply Euler’s first substitution that is $\,\sqrt{t^2-1}=-t+u\,$ and obtain that

$\displaystyle\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt=\int\frac{-2}{(u-1)^2}\,\mathrm du=\frac2{u-1}+C\;.$

Hence ,

$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\frac2{x^2-1+\sqrt{x^4-1}}+C$

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Addendum:

Note that

$\begin{align}\dfrac2{x^2-1+\sqrt{x^4-1}}+C&= \dfrac2{x^2-1+\sqrt{x^4-1}}+1+C-1=\\&=\dfrac{x^2+1+\sqrt{x^4-1}}{x^2-1+\sqrt{x^4-1}}+C-1=\\&=\dfrac{\sqrt{x^2+1}\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}{\sqrt{x^2-1}\left(\sqrt{x^2-1}+\sqrt{x^2+1}\right)}+C-1=\\&=\dfrac{\sqrt{x^2+1}}{\sqrt{x^2-1}}+C-1=\\&=\sqrt{\dfrac{x^2+1}{x^2-1}}+C^*\,.\end{align}$