7
$\begingroup$

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

I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$ and differentiating both sides w.r.t.$x$ as $$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$ Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$

The question is: How to do it otherswise?

$\endgroup$
2
  • 4
    $\begingroup$ This is perfect and then $\to +1$ ! Why do you want to make life difficult ? In this case, it works. It will not be the case very often. $\endgroup$ Commented Jul 25, 2020 at 6:09
  • 1
    $\begingroup$ Yes, tI is like a guess work which has worked thankfully, so the question of other method arises. $\endgroup$ Commented Jul 25, 2020 at 6:13

4 Answers 4

6
$\begingroup$

$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{2x^{-3}-1}{(x^{-2}+x)^{3/2}} dx$$ Now substitute $t=x^{-2}+x$.

$\endgroup$
1
  • 1
    $\begingroup$ This is really elegant! I'll have to save this for the next integration bee $\endgroup$ Commented Jul 25, 2020 at 6:55
4
$\begingroup$

Split the integral and integrate by parts:

$$\begin{align}I = \int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx &\equiv\int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \int\dfrac{x^2}{\left(1 + x^3\right)^{3/2}}x\,\mathrm dx \\ &= \int\dfrac2{\left(1 + x^3\right)^{3/2}}\,\mathrm dx - \left(-\dfrac{2x}{3\left(1 + x^3\right)^{1/2}} + \dfrac23\int\dfrac{1 + x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx\right) \\ &= \int\dfrac{6 - 2 - 2x^3}{3\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23\int\dfrac{2 - x^3}{\left(1 + x^3\right)^{3/2}}\,\mathrm dx + \dfrac{2x}{3\sqrt{1 + x^3}} \\ &= \dfrac23I + \dfrac{2x}{3\sqrt{1 + x^3}} \\ \implies I &= \dfrac{2x}{\sqrt{1 + x^3}}\end{align}$$

$\endgroup$
3
$\begingroup$

@ClaudeLeibovici has a point, because what you did is a well-worn technique, that of using an Ansatz. The basic idea is to make an educated guess as to the form of the solution, then make it more specific with your calculations, as you did. So it's worth understanding what makes a specific Ansatz a sensible starting point:

  • It makes sense to assume a $(1+x^2)^{-3/2}$ factor results from differentiating $(1+x^2)^{-1/2}$: if nothing else, integration by parts makes sense of that.
  • You could have started with a more general Ansatz, $I(x)=f(x)(1+x^3)^{-3/2}$, so the problem is equivalent to $(x^3+1)f^\prime-\tfrac32x^2f=2-x^3$. Since constant $f$ doesn't solve this, it's natural to try a linear $f$ next, which worked for you.

Maybe this answer isn't what you were looking for, but it's important to understand how to use an Ansatz as more than an accident.

$\endgroup$
2
$\begingroup$

$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{\frac{1}{x^3}(2-x^3)}{\frac1{x^3}\left(1+x^3\right)^{3/2}} \ dx$$ $$=\int \frac{\left(\frac{2}{x^3}-1\right)dx}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$ $$=-\int \frac{d\left(\frac{1}{x^2}+x\right)}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$ $$=- \frac{\left(\frac{1}{x^2}+x\right)^{-\frac32+1}}{-\frac32+1}+C$$ $$=\frac{2}{\sqrt{\frac1{x^2}+x}}+C$$ $$=\bbox[15px, #ffd, border:1px solid green]{\frac{2x}{\sqrt{1+x^3}}+C}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .