Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{{{x}^{2}}}{1-{{x}^{2}}} \right)$.[1] Hence find \

Question

Find $\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{{{x}^{2}}}{1-{{x}^{2}}} \right)$.[1] Hence find $\displaystyle\int \frac{2x\ln x}{{{\left( 1-{{x}^{2}} \right)}^{2}}}\,\mathrm{d}x$.[3] Using the substitution $x=\sin \theta $, where $ 0 < \theta < \frac{\pi }{2} $, show that $8\displaystyle\int{{{x}^{2}}\sqrt{1-{{x}^{2}}}\,\mathrm{d}x={{\sin }^{-1}}x-\mathrm{p}(x)\sqrt{1-{{x}^{2}}}+C}$, where $\mathrm{p}(x)$ is a cubic polynomial to be determined.[5]

Accepted Answer

(a)(i) $\frac{2x}{(1-x^2)^2}$ (a)(ii) $\frac{x^2 \ln x}{1-x^2} + \frac{1}{2} \ln|1-x^2| + C$ (b) $p(x) = x(1-2x^2)$

2023 ASRJC Promo Q8

2023 ASRJC Promo Q8

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