$\int \sqrt{\frac{1-{x}}{1+{x}}.\frac{1}{x^2}}dx$

Question

solve: $$\int \sqrt{\frac{1-{x}}{1+{x}}.\frac{1}{x^2}} dx$$

=$$\int \sqrt\frac{1-{x}}{{x^2}+{x^3}} dx$$ I dont know what to do ...

score 1 · Answer 1 · answered Oct 18 '19 at 14:07

1

Hint

Let $\sqrt{\dfrac{1-x}{1+x}}=t$

$x=\dfrac{1-t^2}{1+t^2}=2-\dfrac1{1+t^2}$

$dx=?$

$$I=\int\dfrac{2t^2}{(1+t^2)|1-t^2|}dt$$

Now use partial fraction

answered Oct 18 '19 at 14:07

lab bhattacharjee

How can I recognize that I should use this substitution? $\sqrt{\dfrac{1-x}{1+x}}=t$ – User Oct 18 '19 at 14:20
1

@Soheil, Rationalization often works fine – lab bhattacharjee Oct 18 '19 at 14:21

1 Answers1