Find \(\displaystyle\displaystyle \displaystyle\int{{{x}^{2}}}{{\tan }^{-1}}\left( 2x \right)\,\,\ma

Question

Find $\displaystyle\displaystyle \displaystyle\int{{{x}^{2}}}{{\tan }^{-1}}\left( 2x \right)\,\,\mathrm{d}x$.[3] Find $\displaystyle\displaystyle \displaystyle\int{\frac{x}{\sqrt{8-2x-{{x}^{2}}}}\mathrm{d}x}$.[3]

Accepted Answer

(a) $\int x^2\tan^{-1}(2x)dx=\frac13x^3\tan^{-1}(2x)-\frac1{12}x^2+\frac1{48}\ln|1+4x^2|+C$ (b) $\int\frac{x}{\sqrt{8-2x-x^2}}dx=-\sqrt{8-2x-x^2}-\sin^{-1}\left(\frac{x+1}{3}\right)+C$

2021 ASRJC Promo Q3

2021 ASRJC Promo Q3

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