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Question Number 96864 by bemath last updated on 05/Jun/20

$$\int \frac{\mathrm{dy}}{{\mathrm{y}}^2(5-{\mathrm{y}}^2)}?$$

Commented by bemath last updated on 05/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by Sourav mridha last updated on 05/Jun/20

$$=\frac{1}{5}[\int \frac{\mathit{d}\mathit{y}}{{\mathit{y}}^2}+\int \frac{\mathit{d}\mathit{y}}{{\left(\sqrt{5}\right)}^2-{\mathit{y}}^2}]$$$$=-\frac{1}{5\mathit{y}}+\frac{1}{10\sqrt{5}}\mathit{l}\mathit{n}\left[\frac{\sqrt{5}+\mathit{y}}{\sqrt{5}-\mathit{y}}\right]+\mathit{c}$$

Answered by mathmax by abdo last updated on 05/Jun/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2(5-{\mathrm{x}}^2)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2(5-{\mathrm{x}}^2)}=\frac{1}{{\mathrm{x}}^2(\sqrt{5}-\mathrm{x})(\sqrt{5}+\mathrm{x})}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{{\mathrm{x}}^2}+\frac{\mathrm{c}}{\sqrt{5}-\mathrm{x}}+\frac{\mathrm{d}}{\sqrt{5}+\mathrm{x}}$$$$\mathrm{b}=\frac{1}{5},\mathrm{c}=\frac{1}{5\left(2\sqrt{5}\right)}=\frac{1}{10\sqrt{5}},\mathrm{d}=\frac{1}{5\left(2\sqrt{5}\right)}=\frac{1}{10\sqrt{5}}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{1}{{5\mathrm{x}}^2}+\frac{1}{10\sqrt{5}(\sqrt{5}-\mathrm{x})}+\frac{1}{10\sqrt{5}(\sqrt{5}+\mathrm{x})}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}-\mathrm{c}+\mathrm{d}\Rightarrow \mathrm{a}=\mathrm{c}-\mathrm{d}=0\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{5\mathrm{x}}^2}+\frac{1}{10\sqrt{5}(\sqrt{5}-\mathrm{x})}+\frac{1}{10\sqrt{5}(\sqrt{5}+\mathrm{x})}\Rightarrow$$$$\mathrm{I}=-\frac{1}{5\mathrm{x}}-\frac{1}{10\sqrt{5}}\ln \mid \mathrm{x}-\sqrt{5}\mid +\frac{1}{10\sqrt{5}}\ln \mid \mathrm{x}+\sqrt{5}\mid +\mathrm{C}$$$$=-\frac{1}{5\mathrm{x}}+\frac{1}{10\sqrt{5}}\ln \mid \frac{\mathrm{x}+\sqrt{5}}{\mathrm{x}-\sqrt{5}}\mid +\mathrm{C}$$

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