Find $\displaystyle\displaystyle \displaystyle\int{\frac{1-3x}{1+9{{x}^{2}}}}\,\mathrm{d}x$.[3] Us

Question

Find $\displaystyle\displaystyle \displaystyle\int{\frac{1-3x}{1+9{{x}^{2}}}}\,\mathrm{d}x$.[3] Use the substitution $x=2\sin \theta $, where $\theta $ is acute, to find $\displaystyle\displaystyle \displaystyle\int{\frac{{{\left( x-1 \right)}^{2}}}{\sqrt{4-{{x}^{2}}}}\mathrm{d}x}$.[5]

Accepted Answer

(a) $\frac{1}{3}{{	an }^{-1}}\left( 3x ight)-\frac{1}{6}\ln \left| 1+9{{x}^{2}} ight|+c$ (b) $3{{\sin }^{-1}}\left( \frac{x}{2} ight)-\frac{1}{2}x\sqrt{4-{{x}^{2}}}+2\sqrt{4-{{x}^{2}}}+c$

2022 DHS Promo Q6

2022 DHS Promo Q6

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