Jtdylktuy

How do I start with evaluating this-

$$intfrac{1+x^4}{(1-x^4)^{3/2}}dx$$

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.

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

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}{2sqrt{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}{2sqrt{1-x^4}}
end{align}
for $x in (-1,1)$ . Now we can solve this equation for your integral.

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'=afrac{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 $I=intfrac{1+x^2}{(1-x^2)^{2}}dx$ and let $f(x)=x^{-1}-x$, $f'(x)=-x^{-2}-1=-1cdot(x^{-2}+1)$, $a=-1$. Now by $(1)$
$$frac{d,[f(x)]^{-1}}{dx}=-1cdotfrac{(-1cdot(x^{-2}+1))}{(x^{-1}-x)^{1-(-1)}}
=frac{(x^{-2}+1)}{(x^{-1}-x)^{2}}=frac{1+x^2}{(1-x^2)^2}$$
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 $f(x)=x^{-2}-x^{2}$, $f'(x)=-2x^{-3}-2x=-2cdot(x^{-3}+x)$, $a=-frac{1}{2}$. Now by $(1)$
$$frac{d,[f(x)]^{-1/2}}{dx}=-frac{1}{2}cdotfrac{(-2cdot(x^{-3}+x))}{(x^{-2}-x^{2})^{1-(-frac{1}{2})}}
=frac{(x^{-3}+x)}{(x^{-2}-x^{2})^{3/2}}=frac{1+x^4}{(1-x^4)^{3/2}}$$
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&=intfrac{1+x^6}{(1-x^6)^{4/3}}dx=intfrac{(1+x^6)/x^4}{(1-x^6)^{4/3}/x^4}dx=intfrac{x^{-4}+x^2}{(x^{-3}-x^3)^{4/3}}dx\
&=int dleft((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 $f(x)=x^{-n}-x^{n}$, $f'(x)=-ncdot(x^{-(n+1)}+x^{n-1})$, $a=-frac{1}{n}$)
begin{align*}
I&=intfrac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=intfrac{(1+x^{2n})/x^{(n+1)}}{(1-x^{2n})^{(n+1)/n}/x^{(n+1)}}dx=int-frac{1}{n}cdotfrac{left(-ncdot(x^{-{(n+1)}}+x^{n-1})right)}{(x^{-n}-x^{n})^{(n+1)/n}}dx\
&=int dleft((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=intfrac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=frac{x}{(1-x^{2n})^{1/n}}+c
end{equation}

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

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

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