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I have to find the integral of $\frac{1}{(x+1)\sqrt{x^2+2x}}$. So I thought about writing it as $\frac{1}{(x+1)\sqrt{(x+1)^2-1}}$ here I replace $x+1=u$ and I have $\frac{1}{u(u^2-1)}$ . What do I do next? The answer on my textbook is $-\arcsin[\frac{1}{x+1}]$.

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Put $x+1=\sec\theta\implies x^2+2x=(x+1)^2-1=\tan^2\theta$ –  lab bhattacharjee Jan 24 '13 at 17:51
    
You forgot the square root in the denominator after the substitution. I'm pretty sure I saw this same problem recently. In any case, depending on how you do this, you may end up with several different correct forms. –  Mike Jan 24 '13 at 18:43

4 Answers 4

You can do the integral by using the $\sqrt{x^2+2x}=t- x$ as well.

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Nice alternative! –  amWhy Feb 6 '13 at 2:58

Hint: substitute $u = \sec{y}$, $du = \sec{y} \tan{y} dy$. Then $u^2-1 = \tan^2{y}$.

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Your substitution is good. Now one can use a trigonometric substitution. But the substitution $u=\frac{1}{t}$ works very nicely. Then $du=-\frac{dt}{t^2}$, and after some minor algebra we arrive at the very familiar $$\int -\frac{dt}{\sqrt{1-t^2}},$$ and we are one step from the answer you mentioned.

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Let $x+1=u$ and then $u=1/t$ $$\int \frac{du}{u\sqrt{u^2-1}}=-\int \frac{dt}{\sqrt{1-t^2}}=-\arcsin t +C$$ Thus $$\int\frac{1}{(x+1)\sqrt{x^2+2x}}\mathrm{dx}=-\arcsin\left(\frac{1}{x+1}\right)+C$$

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What I noted below as a substitution, will lead the integral to an integration of quotient function which is less beautiful. Yours is much practical and the Andre's also. +1 –  B. S. Jan 24 '13 at 18:41
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@BabakSorouh: thank you! I'm glad you like my answers! :-) –  Chris's sis Jan 24 '13 at 18:42

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