Mathematics

# $\displaystyle \int\{\frac{(\log x-1)}{1+(\log x)^{2}}\}^{2}dx$ is equal to

$\displaystyle \frac{x}{(\log x)^{2}+1}+c$

##### SOLUTION
$\int [\dfrac {log x-1}{1+(log x)^2}]^2dx.$

$\int \dfrac {(log x-1)^2}{[1+(log x)^2]^2}=\int \dfrac {(log x)^2+1-2 log x.)}{[1-(log x)^2]^2.}$

$=\int (\dfrac {1}{[(1+(log x)^2]}-\dfrac {2 log x}{[1+log^2x]^2})dx$

$=\int \dfrac {1}{(1+log^2x)}dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx.$

$=\dfrac {1}{(1+log^2x)}\int 1 dx+\int [\dfrac {2 log x}{(1+log^2x)^2}-\dfrac {1}{x}\int 1\cdot dx]dx\cdot-\int \dfrac {2 log x}{(1+log^2x)^2}dx$

$=\dfrac {x}{1+log^2x}+\int \dfrac {2 log x}{(1+log^2x)^2}dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx+c$

$=\dfrac {x}{1+log^2x}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 113

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