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  4. ∫ (log x-1/1+(log x)2+1) 2 dx is equal to

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Q. $\int\left\{\frac{log \,x-1}{1+\left(log \,x\right)^{2}+1}\right\}^{2}$ dx is equal to

Integrals

Solution:

$\int\frac{\left(log \,x-1\right)^{2}}{\left(1+\left(log\, x\right)^{2}\right)^{2}}dx$
$=\int\left[\frac{1}{\left(1+\left(log \,x\right)\right)^{2}}-\frac{2\,log \,x}{\left(1+\left(log\, x\right)^{2}\right)^{2}}\right]dx$
$=\int\left[\frac{e^{t}}{1+t^{2}}-\frac{2t e^{t}}{\left(1+t^{2}\right)^{2}}\right]dt$
( Putting $log\, x = t \Rightarrow dx =e^{t} dt)$
$=\int e^{t}\left[\frac{1}{1+t^{2}}-\frac{2t}{\left(1+t^{2}\right)^{2}}\right]dt$
$=\frac{e^{t}}{1+t^{2}}+c$
$=\frac{x}{1+\left(log \,x\right)^{2}}+c \,\,$(Resubstituting $t = log\, x)$