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:-

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

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

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

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

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

if $t = 1-x^4$ shouldn't the numerator be $(1-t) + (1-t)^4$? — qwr, Dec 26 '19 at 01:27
"Solve" is the wrong word here. One solves equations; one solves problems. One evaluates expressions, and that is what one has here. — Michael Hardy, Dec 26 '19 at 04:32

Simply Beautiful Art · Answer 1 · 2019-12-26T01:32:15.127

7

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:

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

edited Dec 26 '19 at 01:32

answered Dec 26 '19 at 01:26

Simply Beautiful Art

74,685

power in denominator is $\frac{7}{2}$, not $\frac{5}{2}$ – user3290550 Dec 26 '19 at 01:31
Sorry, typos.${}$ – Simply Beautiful Art Dec 26 '19 at 01:32
can you correct your latex? – user3290550 Dec 26 '19 at 01:33
@Ducan, what is the advantage? – user3290550 Dec 26 '19 at 01:35
how did you get the idea of differentiating $f(x)$ with $(1-x^4)^{\frac{5}{2}}$ – user3290550 Dec 26 '19 at 01:41
It is apparent that when differentiating such a rational function, the "degree" of the denominator increases by 1. – Simply Beautiful Art Dec 26 '19 at 01:42
but if we are expected to get the result by solving the integration, then? – user3290550 Dec 26 '19 at 01:57
@user3290550 Integration by parts combined with chain rule, where you handle a part of the integral using $u=(1-x^4)^{5/2}$. In essence, reversed quotient rule (which is product rule and chain rule applied to $f(x)[g(x)]^{-1}$.) – Simply Beautiful Art Dec 26 '19 at 02:07

Quanto · Answer 2 · 2023-05-11T23:51:29.293

2

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

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

edited May 11 '23 at 23:51

answered Dec 26 '19 at 03:01

Quanto

97,352

Intriguing approach. My bible is Calculus (volumes 1 and 2) by Apostol, which does not contain any guidance for adopting your approach as a first try (i.e. it does, for example, advocate attacking $\int R(x, \sqrt{a^2 - [cx + d]^2})dx$ via $cx+d = a \sin t$. Is there a calculus textbook that explores problems similar to the one queried here, advocating customized approaches for each such problem? – user2661923 Dec 26 '19 at 04:47
1

Integrals such as this one barely have any practical significance. They are customarily designed to explore ‘perfect’ substitution identified here. They are many of kinds for which the textbook techniques for integration are ineffective. Another example is $\int \frac1{(1+x)\sqrt{1-x^2}}dx=-2\sqrt t$, where the perfect substitute is $t=\frac{1-x}{1+x}$ – Quanto Dec 26 '19 at 05:30

lab bhattacharjee · Answer 3 · 2019-12-28T11:46:27.890

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 $n=2\iff2m=7$

First of all $$m\ne-1$$ and we can different values of $n$ to generate many such integral problems. For example set $2m+1=4\cdot3$ — lab bhattacharjee, Dec 28 '19 at 12:02

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

3 Answers3