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Question Number 108208 by bemath last updated on 15/Aug/20

$$\frac{\mathcal{B}e\mathcal{M}ath}{\boxminus }$$$$\left(1\right)\int \frac{x^5-x}{x^8+1}dx$$$$\left(2\right)\underset{\frac{1}{\sqrt{2}}}{\stackrel{1}{\int }}\frac{{\sin }^{-1}\left(x\right)}{x^3}dx$$

Answered by john santu last updated on 15/Aug/20

$$\frac{★\mathcal{JS}\mathrm{♠}}{\Bumpeq }$$$$J=\int \frac{x(x^4-1)}{x^8+1}dx$$$$letz=x^2\Rightarrow J=\frac{1}{2}\int \frac{z^2-1}{z^4+1}dz$$$$decomposition$$$$z^4+1={(z^2+1)}^2-2z^2$$$$=(z^2+z\sqrt{2}+1)(z^2-z\sqrt{2}+1)$$$$\frac{z^2-1}{z^4+1}=\frac{1}{2}(\frac{z\sqrt{2}-1}{z^2-z\sqrt{2}+1}-\frac{z\sqrt{2}+1}{z^2+z\sqrt{2}+1})$$$$J=\frac{1}{4}\int (\frac{z\sqrt{2}-1}{z^2-z\sqrt{2}+1}-\frac{z\sqrt{2}+1}{z^2+z\sqrt{2}+1})dz$$$$J=\frac{1}{4\sqrt{2}}\ln \left(\frac{x^4-x^2\sqrt{2}+1}{x^4+x^2\sqrt{2}+1}\right)+c$$

Commented by bobhans last updated on 15/Aug/20

$$\frac{\mathcal{C}\mathrm{ooll}}{\mathrm{great}}$$

Answered by bobhans last updated on 15/Aug/20

$$\left(2\right)\frac{\mathrm{bobhans}}{°\bullet \equiv \bullet °}$$$$\mathrm{I}=\underset{1/\sqrt{2}}{\stackrel{1}{\int }}\frac{{\sin }^{-1}\left(\mathrm{x}\right)}{{\mathrm{x}}^3}\mathrm{dx}$$$$\mathrm{by}\mathrm{part}\{\begin{array}{l}\mathrm{u}={\sin }^{-1}\left(\mathrm{x}\right)\to \mathrm{du}=\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}\\ \mathrm{v}=-\frac{1}{{2\mathrm{x}}^2}\end{array}$$$$\mathrm{I}={[-\frac{{\sin }^{-1}\left(\mathrm{x}\right)}{{2\mathrm{x}}^2}]}_{\frac{1}{\sqrt{2}}}^1+\underset{1/\sqrt{2}}{\stackrel{1}{\int }}\frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1-{\mathrm{x}}^2}}$$$$\mathrm{I}=0+\frac{1}{2}\underset{\pi /4}{\stackrel{\pi /2}{\int }}\frac{\cos \mathrm{h}\mathrm{dh}}{\sin {}^2\mathrm{h}\sqrt{1-\sin {}^2\mathrm{h}}}$$$$\mathrm{I}=\frac{1}{2}\underset{\pi /4}{\stackrel{\pi /2}{\int }}{\csc }^2\mathrm{h}\mathrm{dh}=-\frac{1}{2}{\left[\cot \mathrm{h}\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$$$$\mathrm{I}=\frac{1}{2}$$

Answered by mathmax by abdo last updated on 15/Aug/20

$$2)\mathrm{I}={\int }_{\frac{1}{\sqrt{2}}}^1\frac{\mathrm{arcsinx}}{{\mathrm{x}}^3}\mathrm{dx}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}={\mathrm{x}}^{-3}\mathrm{and}\mathrm{v}=\mathrm{arcsinx}$$$$\Rightarrow \mathrm{I}={[-\frac{1}{2}{\mathrm{x}}^{-2}\mathrm{arcsinx}]}_{\frac{1}{\sqrt{2}}}^1+\frac{1}{2}{\int }_{\frac{1}{\sqrt{2}}}^1{\mathrm{x}}^{-2}\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}$$$$=\frac{\pi }{4}-\frac{\pi }{4}+\frac{1}{2}{\int }_{\frac{1}{\sqrt{2}}}^1\frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1-{\mathrm{x}}^2}}=\frac{1}{2}{\int }_{\frac{1}{\sqrt{2}}}^1\frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1-{\mathrm{x}}^2}}$$$${=}_{\mathrm{x}=\mathrm{sint}}\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{\mathrm{cost}\mathrm{dt}}{{\sin }^2\mathrm{t}\mathrm{cost}}=\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{2\mathrm{dt}}{1-\cos \left(2\mathrm{t}\right)}$$$${=}_{2\mathrm{t}=\mathrm{u}}{\int }_{\frac{\pi }{2}}^{\pi }\frac{\mathrm{du}}{2\{1-\mathrm{cosu}\}}{=}_{\tan \left(\frac{\mathrm{u}}{2}\right)=\mathrm{z}}\frac{1}{2}{\int }_1^{+\mathrm{\infty }}\frac{2\mathrm{dz}}{(1+{\mathrm{z}}^2)(1-\frac{1-{\mathrm{z}}^2}{1+{\mathrm{z}}^2})}$$$$={\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dz}}{1+{\mathrm{z}}^2-1+{\mathrm{z}}^2}=\frac{1}{2}{\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dz}}{{\mathrm{z}}^2}=\frac{1}{2}{[-\frac{1}{\mathrm{z}}]}_1^{+\mathrm{\infty }}=\frac{1}{2}\Rightarrow$$$$\mathrm{I}=\frac{1}{2}$$$$$$

Answered by mathmax by abdo last updated on 15/Aug/20

$$1)\mathrm{complex}\mathrm{method}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^5-\mathrm{x}}{{\mathrm{x}}^8+1}$$$${\mathrm{x}}^8+1=0\Rightarrow {\mathrm{x}}^8={\mathrm{e}}^{\mathrm{i}(2\mathrm{k}+1)\pi }\Rightarrow {\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\frac{\mathrm{i}(2\mathrm{k}+1)\pi }{8}}\mathrm{and}\mathrm{k}\in \left[[0,7]\right]$$$$\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^5-\mathrm{x}}{\prod _{\mathrm{k}=0}^7(\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})}=\sum _{\mathrm{k}=0}^7\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}$$$${\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{z}}_{\mathrm{k}}^5-{\mathrm{z}}_{\mathrm{k}}}{{8\mathrm{z}}_{\mathrm{k}}^7}=\frac{{\mathrm{z}}_{\mathrm{k}}^6-{\mathrm{z}}_{\mathrm{k}}^2}{-8}\Rightarrow \int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{8}\int \sum _{\mathrm{k}=0}^7\frac{{\mathrm{z}}_{\mathrm{k}}^6-{\mathrm{z}}_{\mathrm{k}}^2}{\mathrm{x}-{\mathrm{z}}_{\mathrm{k}}}\mathrm{dx}$$$$=-\frac{1}{8}\sum _{\mathrm{k}=0}^7({\mathrm{z}}_{\mathrm{k}}^6-{\mathrm{z}}_{\mathrm{k}}^2)\ln (\mathrm{x}-{\mathrm{z}}_{\mathrm{k}})+\mathrm{C}$$

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