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Question Number 106705 by bemath last updated on 06/Aug/20

$$@\mathrm{bemath}@$$$$\int \frac{1+{\mathrm{x}}^2}{\sqrt{9-{4\mathrm{x}}^2}}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 06/Aug/20

$$\mathrm{I}=\int \frac{1+{\mathrm{x}}^2}{\sqrt{9-{4\mathrm{x}}^2}}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{1+{\mathrm{x}}^2}{3\sqrt{1-\frac{4}{9}{\mathrm{x}}^2}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$\frac{2}{3}\mathrm{x}=\mathrm{sint}\Rightarrow \mathrm{I}=\int \frac{1+\frac{9}{4}{\sin }^2\mathrm{t}}{3\mathrm{cost}}\times \frac{3}{2}\mathrm{cost}\mathrm{dt}$$$$=\frac{1}{2}\int \frac{4+{9\sin }^2\mathrm{t}}{4}\mathrm{dt}=\frac{1}{2}\mathrm{t}+\frac{9}{8}\int {\sin }^2\mathrm{t}\mathrm{dt}$$$$=\frac{\mathrm{t}}{2}+\frac{9}{16}\int (1-\cos \left(2\mathrm{t}\right))\mathrm{dt}=\frac{\mathrm{t}}{2}+\frac{9\mathrm{t}}{16}-\frac{9}{32}\sin \left(2\mathrm{t}\right)+\mathrm{C}$$$$=\frac{17\mathrm{t}}{16}-\frac{9}{16}\mathrm{sint}\mathrm{cost}+\mathrm{C}$$$$=\frac{17}{16}\arcsin \left(\frac{2\mathrm{x}}{3}\right)-\frac{9}{16}.\frac{2\mathrm{x}}{3}\sqrt{1-\frac{4}{9}{\mathrm{x}}^2}+\mathrm{C}$$$$\mathrm{I}=\frac{17}{16}\arcsin \left(\frac{2\mathrm{x}}{3}\right)-\frac{3\mathrm{x}}{8}\sqrt{1-\frac{4}{9}{\mathrm{x}}^2}+\mathrm{C}$$

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