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How can I calculate the following integral :

$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$

I try to write the integral like :

$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{(x+1)}$$

but still nothing .

thanks :)

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You can use the Euler substitution $\sqrt{1+x^2}=t-x$ to reduce the integration of a rational of $x$ and $\sqrt{1+x^2}$ to an integration of a rational fraction in $t$. –  Américo Tavares Jan 15 '13 at 21:01
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2 Answers 2

up vote 3 down vote accepted

Let $x=\tan[y]$, then $dx=\sec^{2}[y]dy$ $$\frac{1}{1+x}=\frac{\cos[y]}{\sin[y]+\cos[y]},\frac{1}{\sqrt{1+x^{2}}}=\frac{1}{\sec[y]}=\cos[y]$$

Multiplying everything out you need to integrate $$\frac{1}{\sin[y]+\cos[y]}dy$$

But you can simplify $$\sin[y]+\cos[y]=\sqrt{2}[\sin[y]\cos[\frac{\pi}{4}]+\cos[y]\sin[\frac{\pi}{4}]]=\sqrt{2}\sin[y+\frac{\pi}{4}]$$

So it suffice to integrate $$\frac{1}{\sqrt{2}}\csc[y+\frac{\pi}{4}]dy$$And we know how to do $\int \csc[x]dx$.

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Try to substitute $x=\tan t$ and after some manipulations recall that $$ \left(\ln \tan \frac{u}{2}\right)'= \frac{u'}{\sin u} $$

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