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

Mathematics Asked by Dharmendra Singh on August 5, 2020

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?

4 Answers

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

EDIT: Always try to manipulate such integrands like this. Start by taking out the highest power from the denominator. If that doesn't work, then move on to taking out lesser and lesser powers. If all endeavors fail, resort to good old techniques.

Correct answer by Sameer Baheti on August 5, 2020

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

Answered by Harish Chandra Rajpoot on August 5, 2020

Split the integral and integrate by parts:

$$begin{align}I = intdfrac{2 - x^3}{left(1 + x^3right)^{3/2}},mathrm dx &equivintdfrac2{left(1 + x^3right)^{3/2}},mathrm dx - intdfrac{x^2}{left(1 + x^3right)^{3/2}}x,mathrm dx \ &= intdfrac2{left(1 + x^3right)^{3/2}},mathrm dx - left(-dfrac{2x}{3left(1 + x^3right)^{1/2}} + dfrac23intdfrac{1 + x^3}{left(1 + x^3right)^{3/2}},mathrm dxright) \ &= intdfrac{6 - 2 - 2x^3}{3left(1 + x^3right)^{3/2}},mathrm dx + dfrac{2x}{3sqrt{1 + x^3}} \ &= dfrac23intdfrac{2 - x^3}{left(1 + x^3right)^{3/2}},mathrm dx + dfrac{2x}{3sqrt{1 + x^3}} \ &= dfrac23I + dfrac{2x}{3sqrt{1 + x^3}} \ implies I &= dfrac{2x}{sqrt{1 + x^3}}end{align}$$

Answered by an4s on August 5, 2020

@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.

Answered by J.G. on August 5, 2020

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