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Problem : Solve : $\int \frac{dx}{(x+\sqrt{(x^2-1)})^2}$......(i)

I tried :

Let $x =\sec\theta$ therefore , (i) will become after some simplification

$$\int \frac{\sin\theta}{(1+\sin\theta)^2}d\theta$$

but i think its wrong method of approaching please suggest further....thanks..

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up vote 4 down vote accepted

HINT: $$\frac1{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}$$

So, $$\frac1{(x+\sqrt{x^2-1})^2}=(x-\sqrt{x^2-1})^2=x^2+x^2-1-2x\sqrt{x^2-1}=2x^2-1-2x\sqrt{x^2-1}$$

Put $x^2-1=y$ for the last part

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@sultan, how about this method? – lab bhattacharjee Jun 18 '13 at 3:15

Better to substitute $x=\cosh{u}$. Then the integral becomes

$$\int du \frac{\sinh{u}}{(\cosh{u}+\sinh{u})^2} = \frac12 \int du\, e^{-2 u} (e^{u}-e^{- u}) = \frac12 \left (\frac13 e^{-3 u}-e^{-u}\right)+C$$

Substituting back, where $e^{-u} = x-\sqrt{x^2-1}$, we get that the integral is

$$\frac16 \left ( x-\sqrt{x^2-1}\right)^3 - \frac12 \left ( x-\sqrt{x^2-1}\right) + C$$

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@sultan: how about THIS method? – Ron Gordon Dec 18 '15 at 15:25

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