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Question Number 198420 by Mingma last updated on 19/Oct/23

Answered by Frix last updated on 19/Oct/23

$$\int \frac{x^2}{{(3+4x-4x^2)}^{\frac{3}{2}}}dx\underset{dx=-\frac{(2x-1)\sqrt{3+4x-4x^2}}{4t}}{\stackrel{t=\frac{2+\sqrt{3+4x-4x^2}}{2x-1}}{=}}$$$$=-\frac{1}{16}\int \frac{{(t^2+4t+1)}^2}{{(t-1)}^2{(t+1)}^2(t^2+1)}dt=$$$$=\int (-\frac{9}{32{(t-1)}^2}-\frac{1}{32{(t+1)}^2}+\frac{1}{4(t^2+1)})dt=$$$$=\frac{9}{32(t-1)}+\frac{1}{32(t+1)}+\frac{{\tan }^{-1}t}{4}=$$$$=\frac{5t+4}{16(t^2-1)}+\frac{{\tan }^{-1}t}{4}=$$$$=\frac{10x+3}{32\sqrt{3+4x-4x^2}}+\frac{{\tan }^{-1}\frac{2+\sqrt{3+4x-4x^2}}{2x-1}}{4}+C$$

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