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Question Number 136968 by Eric002 last updated on 28/Mar/21

$$1)\int \frac{8}{x^2\sqrt{4-x^2}}dx$$$$2)\int \frac{x-6}{2x^2-5x+3}$$

Answered by EDWIN88 last updated on 28/Mar/21

$$\left(1\right)\int \frac{8}{{\mathrm{x}}^3\sqrt{{4\mathrm{x}}^{-2}-1}}\mathrm{dx}=\int \frac{{8\mathrm{x}}^{-3}}{\sqrt{{4\mathrm{x}}^{-2}-1}}\mathrm{dx}$$$$\mathrm{set}{\mathrm{u}}^2={4\mathrm{x}}^{-2}-12\mathrm{u}\mathrm{du}=-{8\mathrm{x}}^{-3}\mathrm{dx}$$$$\mathrm{E}=-2\int \frac{\mathrm{u}}{\mathrm{u}}\mathrm{du}=-2\mathrm{u}+\mathrm{c}$$$$\mathrm{E}=-2\sqrt{{4\mathrm{x}}^{-2}-1}+\mathrm{c}=-2\sqrt{\frac{4-{\mathrm{x}}^2}{{\mathrm{x}}^2}}+\mathrm{c}$$$$\mathrm{E}=-\frac{2\sqrt{4-{\mathrm{x}}^2}}{\mathrm{x}}+\mathrm{c}$$

Answered by EDWIN88 last updated on 28/Mar/21

$$\left(2\right)\mathrm{E}=\int \frac{\mathrm{x}-6}{(2\mathrm{x}+1)(\mathrm{x}-3)}\mathrm{dx}$$$$\mathrm{Partial}\mathrm{fraction}\frac{\mathrm{x}-6}{{2\mathrm{x}}^2-5\mathrm{x}+3}=\frac{\mathrm{p}}{2\mathrm{x}+1}+\frac{\mathrm{q}}{\mathrm{x}-3}$$$$\mathrm{where}\{\begin{array}{l}\mathrm{p}=\left[\frac{\mathrm{x}-6}{\mathrm{x}-3}\right]{}_{\mathrm{x}=-\frac{1}{2}}=\frac{-13}{-7}=\frac{13}{7}\\ \mathrm{q}={\left[\frac{\mathrm{x}-6}{2\mathrm{x}+1}\right]}_{\mathrm{x}=3}=-\frac{3}{7}\end{array}$$$$\mathrm{E}=\frac{13}{7}\int \frac{\mathrm{dx}}{2\mathrm{x}+1}-\frac{3}{7}\int \frac{\mathrm{dx}}{\mathrm{x}-3}$$$$\mathrm{E}=\frac{13}{14}\ln \mid 2\mathrm{x}+1\mid -\frac{3}{7}\ln \mid \mathrm{x}-3\mid +\mathrm{c}$$

Answered by mathmax by abdo last updated on 28/Mar/21

$$1)\mathrm{I}=\int \frac{8}{{\mathrm{x}}^2\sqrt{4-{\mathrm{x}}^2}}\mathrm{dx}\Rightarrow \mathrm{I}{=}_{\mathrm{x}=2\sin \theta }\int \frac{8}{{4\sin }^2\theta .2\cos \theta }\left(2\cos \theta \right)\mathrm{d}\theta$$$$=4\int \frac{\mathrm{d}\theta }{{\sin }^2\theta }=8\int \frac{\mathrm{d}\theta }{1-\cos \left(2\theta \right)}{=}_{2\theta =\mathrm{t}}8\int \frac{\mathrm{dt}}{2(1-\mathrm{cost})}=4\int \frac{\mathrm{dt}}{1-\mathrm{cost}}$$$${=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}4\int \frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(1-\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2})}=8\int \frac{\mathrm{dy}}{1+{\mathrm{y}}^2-1+{\mathrm{y}}^2}=4\int \frac{\mathrm{dy}}{{\mathrm{y}}^2}$$$$=-\frac{4}{\mathrm{y}}+\mathrm{C}=-\frac{4}{\tan \left(\frac{\mathrm{t}}{2}\right)}+\mathrm{C}=-\frac{4}{\tan \left(\theta \right)}+\mathrm{C}=-\frac{4}{\tan \left(\arcsin \left(\frac{\mathrm{x}}{2}\right)\right)}+\mathrm{C}$$

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