1-1-u-2-2-u-5-du-please-

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

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

please \dots

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

u = \cos θ

I = - \int \frac{\sin θ d θ}{\sin^{4} θ \cos^{5} θ} = - \int \frac{d θ}{\sin^{3} θ \cos^{5} θ}

s e e Q 129907 for similar case

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

I = \int \frac{du}{{(1 - u^{2})}^{2} u^{5}}

f (u) = \frac{1}{{(1 - u^{2})}^{2} u^{5}} = \frac{(1 - u^{2}) + u^{2}}{{(1 - u^{2})}^{2} u^{5}} = \frac{1}{(1 - u^{2}) u^{5}} + \frac{1}{{(1 - u^{2})}^{2} u^{3}}

= \frac{(1 - u^{2}) + u^{2}}{(1 - u^{2}) u^{5}} + \frac{(1 - u^{2}) + u^{2}}{{(1 - u^{2})}^{2} u^{3}}

= \frac{1}{u^{5}} + \frac{1}{(1 - u^{2}) u^{3}} + \frac{1}{(1 - u^{2}) u^{3}} + \frac{1}{{(1 - u^{2})}^{2} u}

= \frac{1}{u^{5}} + \frac{(1 - u^{2}) + u^{2}}{(1 - u^{2}) u^{3}} + \frac{(1 - u^{2}) + u^{2}}{(1 - u^{2}) u^{3}} + \frac{(1 - u^{2}) + u^{2}}{{(1 - u^{2})}^{2} u}

= \frac{1}{u^{5}} + \frac{1}{u^{3}} + \frac{1}{(1 - u^{2}) u} + \frac{1}{u^{3}} + \frac{1}{(1 - u^{2}) u} + \frac{1}{(1 - u^{2}) u} + \frac{u}{{(1 - u^{2})}^{2}}

= \frac{1}{u^{5}} + \frac{2}{u^{3}} + 3 (\frac{1}{u} + \frac{u}{1 - u^{2}}) + \frac{u}{{(1 - u^{2})}^{2}}

\int f (u) du = - \frac{1}{{4 u}^{4}} - \frac{1}{u^{2}} + 3 \ln ∣ u ∣ - \frac{3 \ln (1 - u^{2})}{2} + \frac{1}{2 (1 - u^{2})} + C

Commented by stelor last updated on 21/Jan/21

m e r c i \dots thank \dots

Answered by MJS_new last updated on 20/Jan/21

\int \frac{d u}{u^{5} {(1 - u^{2})}^{2}} =

[{Ostrogradski}^{'} s Method]

= - \frac{6 u^{4} - 3 u^{2} - 1}{4 u^{4} (u^{2} - 1)} - 3 \int \frac{d u}{u (u^{2} - 1)} =

= - \frac{6 u^{4} - 3 u^{2} - 1}{4 u^{4} (u^{2} - 1)} - \frac{3}{2} \int \frac{d u}{u - 1} - \frac{3}{2} \int \frac{d u}{u + 1} + 3 \int \frac{d u}{u} =

= - \frac{6 u^{4} - 3 u^{2} - 1}{4 u^{4} (u^{2} - 1)} - \frac{3}{2} \ln ∣ u - 1 ∣ - \frac{3}{2} \ln ∣ u + 1 ∣ + 3 \ln ∣ u ∣ =

= - \frac{6 u^{4} - 3 u^{2} - 1}{4 u^{4} (u^{2} - 1)} + \frac{3}{2} \ln \frac{u^{2}}{∣ u^{2} - 1 ∣} + C

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