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I'm doing some problems that involve identifying the best method of integration and then using that method. I'm given the following: $$\int\frac{dx}{(1-x^{2})^{\frac{3}{2}}} dx$$

I solved it using u-substitution: $$\int\frac{dx}{(1-x^{2})^{\frac{3}{2}}} dx = \int\frac{1}{(1-x^{2})^{\frac{3}{2}}} dx$$ $$u=1-x^2, du = -2x dx$$ $$-\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}du = -\frac{1}{2}\int u^{-\frac{3}{2}}du$$ $$-\frac{1}{2}[-\frac{1}{2}u^{-\frac{1}{2}}] = \frac{1}{4}u^{-\frac{1}{2}} = \frac{1}{4\sqrt{u}}$$ $$= \frac{1}{4\sqrt{(1-x^{2})}} + C$$

My book, however, uses trigonometric substitution to solve this integral and gets an answer of: $$\frac{x}{1-x^{2}}+C$$

Did I do something wrong? Where did I mess up?

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    $\begingroup$ the answer here has pointed out one flaw. Here's another. $$-\frac{1}{2}[-\frac{1}{2}u^{-\frac{1}{2}}] = u^{-\frac{1}{2}}$$. You missed a factor of $\frac14$ $\endgroup$
    – Guy
    Mar 6, 2014 at 17:59

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Your $u$ substitution is wrong

as $\displaystyle 1-x^2=u\implies -2x dx=du, x^2=1-u,x=$sign$(x)\cdot\sqrt{1-u}$

$\displaystyle\implies\int\frac{dx}{(1-x^2)^2}=$sign$\displaystyle(x)\cdot-\int\frac{du}{2\sqrt{1-u}\cdot u^2}$

We can use Partial Fraction Decomposition to get $$\frac1{(1-x^2)^2}=\frac A{1-x}+\frac B{(1-x)^2}+\frac C{1+x}+\frac D{(1+x)^2}$$

Multiply out either sides by $(1-x^2)^2,$ and compare the coefficients the different powers of $x$ to find the values of the arbitrary constants $A,B,C,D$

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  • $\begingroup$ @kwikness, how have you reached at $$-\frac12\int u^{-\frac32}du?$$ where as $$1-x^2=u\implies x^2=1-u\implies x=\sqrt{1-u}$$ $\endgroup$ Mar 6, 2014 at 17:59
  • $\begingroup$ sorry i wrote the problem incorrectly the first couple of lines.. $\endgroup$
    – kwikness
    Mar 6, 2014 at 18:05
  • $\begingroup$ now that i've made corrections to the given equation, can you please tell me where I went wrong? $\endgroup$
    – kwikness
    Mar 6, 2014 at 18:50
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    $\begingroup$ @kwikness, if $$du=-2xdx, dx=-\frac{du}{2x}$$ You have ignored this $x$ $\endgroup$ Mar 6, 2014 at 18:52
  • $\begingroup$ @kwikness, could you solve the latest version of the problem using $u-$ substitution? $\endgroup$ Mar 6, 2014 at 18:59

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