Solve $\int \frac{x^4+x^8}{(1-x^4)^{\frac{7}{2}}} \, dx$

Question

Solve $\int \dfrac{x^4+x^8}{(1-x^4)^{\frac{7}{2}}} \, dx$

My attempt is as follows:-

$$1-x^4=t$$ $$-4x^3=\dfrac{dt}{dx}$$ $$x^3dx=\dfrac{-dt}{4}$$

$$\frac{-1}{4}\cdot\int\dfrac{(1-t)^{\frac{1}{4}}(1-t)}{t^{\frac{7}{2}}}dt$$ $$\frac{-1}{4}\cdot\int\dfrac{(1-t)^\frac{5}{4}}{t^{\frac{7}{2}}}dt$$ $$\frac{-1}{4}\cdot\int\left(\dfrac{1}{t} -1\right)^{\frac{5}{4}}\cdot\left(\dfrac{1}{t} \right)^{\frac{9}{4}}dt$$

$$\frac{-1}{4}\cdot\int\left(\dfrac{1}{t} -1\right)^{\frac{5}{4}}\cdot\left(\dfrac{1}{t} \right)^2\cdot\left(\dfrac{1}{t}\right)^{\frac{1}{4}} \, dt$$

$$\frac{1}{t}-1=y$$ $$\frac{-1}{t^2}=\frac{dy}{dt}$$ $$\frac{dt}{t^2}=-dy$$

$$\frac{1}{4}\int y^\frac{5}{4}(y+1)^{\frac{1}{4}} \, dy$$ $$\frac{1}{4}\int y(y^2+y)^{\frac{1}{4}} \, dy$$ $$\frac{1}{8}\int (2y+1-1)(y^2+y)^{\frac{1}{4}} \, dy$$ $$\frac{1}{8}\left(\int (2y+1)(y^2+y)^{\frac{1}{4}} \, dy-\int (y^2+y)^{\frac{1}{4}} \, dy\right)$$ $$\frac{1}{8}\left(\int (2y+1)(y^2+y)^{\frac{1}{4}} \, dy - \int\left(\left(y+\dfrac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)^\frac{1}{4} \right)$$

How to proceed from here or feel free to suggest some shorter and clean way.

Another way recommended

$$I=I_1+I_2$$ $$I=\int \dfrac{x^4}{(1-x^4)^{\frac{7}{2}}}dx+\int \dfrac{x^8}{(1-x^4)^{\frac{7}{2}}}dx$$

First let's solve $I_2$

$$I_2=\int x^5\left(\dfrac{x^3}{(1-x^4)^{\frac{7}{2}}}\right)dx$$

Integrating by parts:-

$$I_2=\dfrac{1}{10}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}-\dfrac{1}{10}\cdot\int \dfrac{5x^4}{(1-x^4)^{\frac{5}{2}}}dx$$

$$I_2=\dfrac{1}{10}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}-\dfrac{1}{10}\cdot\int \dfrac{5x^4(1-x^4)}{(1-x^4)^{\frac{7}{2}}}dx$$

$$I_2=\dfrac{1}{10}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}-\dfrac{1}{10}\cdot\int \dfrac{5x^4(1-x^4)}{(1-x^4)^{\frac{7}{2}}}dx$$

$$I_2=\dfrac{1}{10}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}-\dfrac{1}{2}\cdot\int \dfrac{x^4}{(1-x^4)^{\frac{7}{2}}}dx+\dfrac{I_2}{2}$$

$$\dfrac{I_2}{2}+\dfrac{I_1}{2}=\dfrac{1}{10}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}$$

$$I_1+I_2=\dfrac{1}{5}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}$$ $$I=\dfrac{1}{5}\cdot\dfrac{x^5}{(1-x^4)^{\frac{5}{2}}}$$

Answer 1

It is often easier to look for what form the antiderivative would be and differentiating that when dealing with messy rational functions.

Use the quotient rule to see that we have:

\begin{align}\frac{\mathrm d}{\mathrm dx}\frac{f(x)}{(1-x^4)^{5/2}}&=\frac{(1-x^4)^{5/2}f'(x)+10x^3(1-x^4)^{3/2}f(x)}{(1-x^4)^5}\\&=\frac{(1-x^4)f'(x)+10x^3f(x)}{(1-x^4)^{7/2}}\end{align}

We seek to then solve

$$x^4+x^8=(1-x^4)f'(x)+10x^3f(x)$$

It is easy to verify that we then have $f(x)=x^5/5$ by substituting in a polynomial of degree $5$ to match the power on the LHS. Hence we conclude:

$$\int\frac{x^4+x^8}{(1-x^4)^{7/2}}~\mathrm dx=\frac{x^5}{5(1-x^4)^{5/2}}+C$$

Answer 2

Let $t=\frac x{\sqrt{1-x^4}}$ and $dt = \frac{1+x^4}{(1-x^4)^{3/2}}dx$

$$\int \dfrac{x^4+x^8}{(1-x^4)^{7/2}} dx =\int t^4dt = \frac15t^5+C$$

Answer 3

As $\displaystyle\int\dfrac{x^3}{(1-x^4)^m}dx=\dfrac1{4(m-1)(1-x^4)^{m-1}},$

$$I(m,n)=\int\dfrac{x^{4n-3}\cdot x^3}{(1-x^4)^m}dx=x^{4n-3}\int\dfrac{x^3}{(1-x^4)^m}dx-\int\left(\dfrac{d(x^{4n-3})}{dx}\int\dfrac{x^3}{(1-x^4)^m}dx\right)dx$$

$$=\dfrac{x^{4n-3}}{4(m-1)(1-x^4)^{m-1}}-(4n-3)\int\dfrac{x^{4n-4}}{4(m-1)(1-x^4)^{m-1}}dx$$

$$=\dfrac{x^{4n-3}}{4(m-1)(1-x^4)^{m-1}}-\dfrac{4n-3}{4(m-1)}\int\dfrac{x^{4n-4}(1-x^4)}{(1-x^4)^m}dx$$

$$\implies I(m,n)\left(1-\dfrac{4n-3}{4(m-1)}\right)+\dfrac{4n-3}{4(m-1)}\cdot I(m,n-1)=\dfrac{x^{4n-3}}{4(m-1)(1-x^4)^{m-1}}$$

Set $1-\dfrac{4n-3}{4(m-1)}=\dfrac{4n-3}{4(m-1)}\iff 4n=2m+1$

$$\implies\dfrac12\left(I\left(m,\dfrac{2m+1}4\right) + I\left(m,\dfrac{2m-3}4\right)\right)=\dfrac{x^{2m+1-3}}{4(m-1)(1-x^4)^{m-1}}$$

Finally set $n=2\iff2m=7$