Mathematics

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


ANSWER

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


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
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