11
$\begingroup$

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.

$\endgroup$
21
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Okay... that's magic! But what led to approach the problem in this way?(+1) $\endgroup$ – tatan Jul 21 '18 at 18:30
  • 5
    $\begingroup$ Well, recently I have spent some time with this integral: $$\int \frac{x^2(\ln x -1)}{x^4-\ln ^4 x} dx$$ and in the end I could do it with the same trick.. $\endgroup$ – Zacky Jul 21 '18 at 18:34
  • 1
    $\begingroup$ I guess the crucial observation here is that $${\left( f^\alpha\right)}' = \alpha\cdot \frac {f'}{{(f)}^{1-\alpha}}.$$ In other words, suppose your denominator is some expression to some weird power. If you can manipulate the expression so as to make the numerator be the derivative of what's being 'powered' in the denominator, you're good. $\endgroup$ – Fimpellizieri Jul 21 '18 at 19:13
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.