Mathematics

The value of $$\overset { 1 }{ \underset { 0 }{  \int}  } \dfrac{8log(1+x)}{1+x^2}dx$$ is:-


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$$\dfrac{\pi}{2}log2$$


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Single Correct Medium Published on 17th 09, 2020
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Let $$\displaystyle f\left ( x \right )=\int \frac{x^{2}dx}{\left ( 1+x^{2} \right )\left ( 1+\sqrt{1+x^{2}} \right )}$$ and $$\displaystyle f\left ( 0 \right )=0.$$ Then $$\displaystyle f\left ( 1 \right )$$ is
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