Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem :

Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$

I tried :

$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$

But its not working....Please guide how to proceed... thanks..

share|cite|improve this question
If you're going to make substitutions, it best to put the $dx$ in explicitly... – DJohnM Jun 18 '13 at 4:43

Hint:when $|x|\lt 1$ $$\int(\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}})dx=\int\frac{dx}{\sqrt{(1-x^2)}} + \frac{-1}{2}\int\frac{-2xdx}{\sqrt{(1-x^2)^3}}$$ take $$u=1-x^2$$

share|cite|improve this answer
thanks I got it .... – sultan Jun 18 '13 at 2:35
@sultan:your welcome – Maisam Hedyelloo Jun 18 '13 at 2:43

What you did is a very good first step. For the second integral, make the substitution $u=1-x^2$. We end up with the easy $\int -\frac{1}{2}u^{-3/2}\,du $.

For the first integral, note that the bottom simplifies to $(1-x^2)\sqrt{1-x^2}$, since the square root is only defined when $|x|\le 1$. There is nice cancellation, and we are integrating $\frac{1}{\sqrt{1-x^2}}$, easy, we get an $\arcsin$.

share|cite|improve this answer

It will work the way you have split the integrand. The first term you can simplify to $\int{\frac{1}{\sqrt{1-x^2}}}dx$ and in the second term you can make $x^2=y$.

share|cite|improve this answer

you had a good guees! see

$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx = \int \frac{1-x^2}{(1-x^2)^{\frac{3}{2}}}+\frac{x}{(1-x^2)^{\frac{3}{2}}}dx =\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx$$ note that $[(1-x^2)^{\frac{-1}{2}}] '= -\frac{1}{2}(1-x^2)^\frac{-3}{2}(-2x)= \frac{x}{(1-x)^{\frac{3}{2}}}$ . Now let's us solve changing the variable $x= \sin t$ then $dx=\cos tdt$ therefore $$\int (1-x^2)^{\frac{-1}{2}}dx=\int (1-\sin^2t)^{\frac{-1}{2}}\cos t \;dt= \int \frac{\cos t}{cost} dt=\int dt=t =\arcsin x$$ So we get $$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx=\frac{x}{(1-x)^{\frac{3}{2}}}+\arcsin x$$

share|cite|improve this answer
I think your second integration needs correcting... – DJohnM Jun 18 '13 at 5:25

Using math_man's substitution throughout, $$x=\sin(u)$$So: $$(1-x^2)^\frac{1}{2}=\cos(u)$$and$$dx=\cos(u) du$$ The integral becomes: $$\int \frac{(\cos^2(u)+\sin(u))}{\cos^3(u)}\cos(u) du=\int du+\int \frac{\sin(u)}{\cos^2(u)}du$$Both these integrals are simple.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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