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Question Number 116844 by bemath last updated on 07/Oct/20

∫ ((8x+sin^(−1) (2x))/( (√(1−4x^2 )))) dx

$$\int\:\frac{\mathrm{8x}+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}\:\mathrm{dx}\: \\ $$

Answered by Dwaipayan Shikari last updated on 07/Oct/20

∫((8x)/( (√(1−4x^2 ))))+∫((sin^(−1) (2x))/( (√(1−4x^2 ))))dx     sin^(−1) 2x=u⇒(2/( (√(1−4x^2 ))))=(du/dx)  −∫(dt/( (√t)))+(1/2)∫udu               (t=1−4x^2  ⇒−8x=(dt/dx))  −2(√t)+(1/4)u^2 +C=−2(√(1−4x^2 ))+(1/4)(sin^(−1) 2x)^2 +C

$$\int\frac{\mathrm{8}{x}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}+\int\frac{{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{dx}\:\:\:\:\:{sin}^{−\mathrm{1}} \mathrm{2}{x}={u}\Rightarrow\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}=\frac{{du}}{{dx}} \\ $$$$−\int\frac{{dt}}{\:\sqrt{{t}}}+\frac{\mathrm{1}}{\mathrm{2}}\int{udu}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({t}=\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \:\Rightarrow−\mathrm{8}{x}=\frac{{dt}}{{dx}}\right) \\ $$$$−\mathrm{2}\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} +{C}=−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\left({sin}^{−\mathrm{1}} \mathrm{2}{x}\right)^{\mathrm{2}} +{C} \\ $$

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