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Question Number 129921 by stelor last updated on 20/Jan/21

$$\int \frac{1}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^5}\mathrm{du}=??$$$$\mathrm{please}...$$

Answered by Ar Brandon last updated on 20/Jan/21

$$\mathrm{u}=\cos \theta$$$$\mathcal{I}=-\int \frac{\sin \theta \mathrm{d}\theta }{{\sin }^4\theta {\cos }^5\theta }=-\int \frac{\mathrm{d}\theta }{{\sin }^3\theta {\cos }^5\theta }$$$$seeQ129907\mathrm{for}\mathrm{similar}\mathrm{case}$$

Answered by Ar Brandon last updated on 20/Jan/21

$$\mathcal{I}=\int \frac{\mathrm{du}}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^5}$$$$\mathrm{f}\left(\mathrm{u}\right)=\frac{1}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^5}=\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^5}=\frac{1}{(1-{\mathrm{u}}^2){\mathrm{u}}^5}+\frac{1}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^3}$$$$=\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{(1-{\mathrm{u}}^2){\mathrm{u}}^5}+\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{{(1-{\mathrm{u}}^2)}^2{\mathrm{u}}^3}$$$$=\frac{1}{{\mathrm{u}}^5}+\frac{1}{(1-{\mathrm{u}}^2){\mathrm{u}}^3}+\frac{1}{(1-{\mathrm{u}}^2){\mathrm{u}}^3}+\frac{1}{{(1-{\mathrm{u}}^2)}^2\mathrm{u}}$$$$=\frac{1}{{\mathrm{u}}^5}+\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{(1-{\mathrm{u}}^2){\mathrm{u}}^3}+\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{(1-{\mathrm{u}}^2){\mathrm{u}}^3}+\frac{(1-{\mathrm{u}}^2)+{\mathrm{u}}^2}{{(1-{\mathrm{u}}^2)}^2\mathrm{u}}$$$$=\frac{1}{{\mathrm{u}}^5}+\frac{1}{{\mathrm{u}}^3}+\frac{1}{(1-{\mathrm{u}}^2)\mathrm{u}}+\frac{1}{{\mathrm{u}}^3}+\frac{1}{(1-{\mathrm{u}}^2)\mathrm{u}}+\frac{1}{(1-{\mathrm{u}}^2)\mathrm{u}}+\frac{\mathrm{u}}{{(1-{\mathrm{u}}^2)}^2}$$$$=\frac{1}{{\mathrm{u}}^5}+\frac{2}{{\mathrm{u}}^3}+3(\frac{1}{\mathrm{u}}+\frac{\mathrm{u}}{1-{\mathrm{u}}^2})+\frac{\mathrm{u}}{{(1-{\mathrm{u}}^2)}^2}$$$$\int \mathrm{f}\left(\mathrm{u}\right)\mathrm{du}=-\frac{1}{{4\mathrm{u}}^4}-\frac{1}{{\mathrm{u}}^2}+3\ln \mid \mathrm{u}\mid -\frac{3\ln (1-{\mathrm{u}}^2)}{2}+\frac{1}{2(1-{\mathrm{u}}^2)}+\mathcal{C}$$

Commented by stelor last updated on 21/Jan/21

$$merc\mathrm{i}...\mathrm{thank}...$$

Answered by MJS_new last updated on 20/Jan/21

$$\int \frac{du}{u^5{(1-u^2)}^2}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=-\frac{6u^4-3u^2-1}{4u^4(u^2-1)}-3\int \frac{du}{u(u^2-1)}=$$$$=-\frac{6u^4-3u^2-1}{4u^4(u^2-1)}-\frac{3}{2}\int \frac{du}{u-1}-\frac{3}{2}\int \frac{du}{u+1}+3\int \frac{du}{u}=$$$$=-\frac{6u^4-3u^2-1}{4u^4(u^2-1)}-\frac{3}{2}\ln \mid u-1\mid -\frac{3}{2}\ln \mid u+1\mid +3\ln \mid u\mid =$$$$=-\frac{6u^4-3u^2-1}{4u^4(u^2-1)}+\frac{3}{2}\ln \frac{u^2}{\mid u^2-1\mid }+C$$

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