Without using a graphing calculator, show that \(\displaystyle\displaystyle \displaystyle\int_{0}^{1

Question

Without using a graphing calculator, show that $\displaystyle\displaystyle \displaystyle\int_{0}^{1}{\frac{{{x}^{3}}}{1+{{x}^{2}}}}\mathrm{ d}x=k\left( 1-\ln 2 \right)$, where $k$ is a real number to be determined.[3] Hence find $\displaystyle\displaystyle \displaystyle\int_{0}^{1}{{{x}^{2}}{{\tan }^{-1}}x}\mathrm{ d}x$, leaving your answer in exact form.[3]

Accepted Answer

$k = \frac{1}{2}$; $\int_0^1 x^2 \tan^{-1} x \, dx = \frac{\pi}{12} - \frac{1}{6}(1-\ln 2)$

2018 TJC Promo Q5

2018 TJC Promo Q5

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