How to evaluate $\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$?

Question

How do I start with evaluating this-

$$\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$$

What should be my first attempt at this kind of a problem where-

• The denominator and numerator are of the same degree
• Denominator involves fractional exponent like $3/2$.

Note:I am proficient with all kinds of basic methods of evaluating integrals.

Answer 1

$$\int\frac{1+x^4}{(1-x^4)^{3/2}}dx=\int\frac{1+x^4}{(x^{-2}-x^2)^{3/2}x^3}dx=\int\frac{x^{-3}+x}{(x^{-2}-x^2)^{3/2}}dx$$ Now guess who's derivative is the numerator?

Answer 2

Here's another approach: If we split the integral and integrate by parts, we find \begin{align} \int \limits_0^x \frac{1+t^4}{(1-t^4)^{3/2}} \, \mathrm{d} t &= \int \limits_0^x \frac{1}{(1-t^4)^{3/2}} \, \mathrm{d} t + \int \limits_0^x \frac{t^3}{(1-t^4)^{3/2}} t \, \mathrm{d} t \\ &= \int \limits_0^x \frac{1}{(1-t^4)^{3/2}} \, \mathrm{d} t + \left[\frac{t}{2\sqrt{1-t^4}}\right]_{t=0}^{t=x} - \frac{1}{2} \int \limits_0^x \frac{1-t^4}{(1-t^4)^{3/2}} \, \mathrm{d} t \\ &= \frac{1}{2} \int \limits_0^x \frac{1+t^4}{(1-t^4)^{3/2}} \, \mathrm{d} t + \frac{x}{2\sqrt{1-x^4}} \end{align} for $x \in (-1,1)$ . Now we can solve this equation for your integral.

Answer 3

Using @Fimpellizieri 's brilliant observation on @Dahaka 's ingenious post: \begin{equation} \frac{d\,[f(x)]^a}{dx}=\frac{d\,f^a}{df}\frac{d\,f}{dx}=af^{a-1}\cdot f'=a\frac{f'}{f^{1-a}}\tag{1} \end{equation} we find the integral in the question belongs to a family that can be found from using $(1)$.

First consider $I=\int\frac{1+x^2}{(1-x^2)^{2}}dx$ and let $f(x)=x^{-1}-x$, $f'(x)=-x^{-2}-1=-1\cdot(x^{-2}+1)$, $a=-1$. Now by $(1)$ $$\frac{d\,[f(x)]^{-1}}{dx}=-1\cdot\frac{(-1\cdot(x^{-2}+1))}{(x^{-1}-x)^{1-(-1)}} =\frac{(x^{-2}+1)}{(x^{-1}-x)^{2}}=\frac{1+x^2}{(1-x^2)^2}$$ and so $$I=\int \frac{(1+x^{2})}{(1-x^{2})^{2}}dx=\int d\,([f(x)]^{-1})=[f(x)]^{-1}+c=(x^{-1}-x)^{-1}+c=\frac{x}{1-x^2}+c$$

For the next let $f(x)=x^{-2}-x^{2}$, $f'(x)=-2x^{-3}-2x=-2\cdot(x^{-3}+x)$, $a=-\frac{1}{2}$. Now by $(1)$ $$\frac{d\,[f(x)]^{-1/2}}{dx}=-\frac{1}{2}\cdot\frac{(-2\cdot(x^{-3}+x))}{(x^{-2}-x^{2})^{1-(-\frac{1}{2})}} =\frac{(x^{-3}+x)}{(x^{-2}-x^{2})^{3/2}}=\frac{1+x^4}{(1-x^4)^{3/2}}$$ and so $$I=\int \frac{1+x^4}{(1-x^4)^{3/2}}dx=\int d\,([f(x)]^{-1/2})=[f(x)]^{-1/2}+c=(x^{-2}-x^{2})^{-1/2}+c=\frac{x}{(1-x^4)^{1/2}}+c$$

For the next in the pattern we have \begin{align*} I&=\int\frac{1+x^6}{(1-x^6)^{4/3}}dx=\int\frac{(1+x^6)/x^4}{(1-x^6)^{4/3}/x^4}dx=\int\frac{x^{-4}+x^2}{(x^{-3}-x^3)^{4/3}}dx\\ &=\int d\left((x^{-3}-x^3)^{-1/3}\right)=(x^{-3}-x^3)^{-1/3}+c=\frac{x}{(1-x^6)^{1/3}}+c \end{align*} In general we have (with $f(x)=x^{-n}-x^{n}$, $f'(x)=-n\cdot(x^{-(n+1)}+x^{n-1})$, $a=-\frac{1}{n}$) \begin{align*} I&=\int\frac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=\int\frac{(1+x^{2n})/x^{(n+1)}}{(1-x^{2n})^{(n+1)/n}/x^{(n+1)}}dx=\int-\frac{1}{n}\cdot\frac{\left(-n\cdot(x^{-{(n+1)}}+x^{n-1})\right)}{(x^{-n}-x^{n})^{(n+1)/n}}dx\\ &=\int d\left((x^{-n}-x^n)^{-1/n}\right)=(x^{-n}-x^n)^{-1/n}+c=\frac{x}{(1-x^{2n})^{1/n}}+c \end{align*} giving the general result: \begin{equation} I=\int\frac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=\frac{x}{(1-x^{2n})^{1/n}}+c \end{equation}

Answer 4

This is an elementary method for begigners. It can be done with simple algebraic manipulation.

$\frac{1+x^4}{(1-x^4)^{3/2}}=\frac{2x^4}{(1-x^4)^{3/2}}+(1-x^4)^{-1/2}= x.[(1-x^4)^{-1/2}]'+ x'.(1-x^4)^{-1/2}=[x(1-x^4)^{-1/2}]'$

$$\int \frac{1+x^4}{(1-x^4)^{3/2}} = x(1-x^4)^{-1/2}+c$$