# Integrate 1 / (x+1)^2 square root (x^2+2x+2)

Solution:

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

$x^{2}+2x+2=(x+1)^{2}+1$

Let  us  put     $x+1=t;$   then

$I=\int \frac{1}{t^{2}\sqrt{t^{2}+1}}dt$

Again we have integral  of the form  I.  make the substitution   $t=\sinh z.$

$dt =\cosh z dz \hspace{1cm}\sqrt{t^{2}+1}=\sqrt{1+\sinh^{2}z}=\cosh z$

Hence ,       $I=\frac{\cosh z dz}{\sinh^{2}z\cosh z}=\int \frac{1}{\sinh^{2}z}dz=-coth z+c$

$I=-\frac{\sqrt{1+t^{2}}}{t}+c=-\frac{\sqrt{x^{2}+2x+2}}{x+1}+c.$

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