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Cant integrate

Solve this integration please!!

Note by Hemant Khatri
3 years, 3 months ago

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I tell you question put up by you is not too hard but way too lengthy. Anyway here's the solution.

\(I=\int { \frac { dx }{ { x }^{ 3 }+{ x }^{ 8 } } } =\int { \frac { dx }{ { x }^{ 3 }(1+{ x }^{ 5 }) } } =\int { \frac { dx }{ { x }^{ 3 } } } -\int { \frac { { x }^{ 2 }dx }{ 1+{ x }^{ 5 } } }\)

Now we can write : \({x}^{5}+1=(x+1)({x}^{2}+2xcos(\frac{2\pi}{5})+1)({x}^{2}-2xcos(\frac{\pi}{5})+1)\)

Hence our \(I\) becomes :

\(I=\frac{-2}{{x}^{2}}-{I}_{1}\)

\({I}_{1}=\int { \frac { { x }^{ 2 }dx }{(x+1)({x}^{2}+2xcos(\frac{2\pi}{5})+1)({x}^{2}-2xcos(\frac{\pi}{5})+1) } }\)

after some tedious applications of partial fractions we get :

\({ I }_{ 1 }=\frac { 1 }{ 5 } (\frac { 1 }{ x+1 } +(\frac { 1}{ 2cos(\frac { \pi }{ 5 } ) } )(\frac { x+1 }{ { x }^{ 2 }-2xcos(\frac { \pi }{ 5 } )+1 } )-(\frac { 1 }{ 2cos(\frac { 2\pi }{ 5 } ) } )(\frac { x+1 }{ { x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1 } ))\)

\(\Rightarrow { I }_{ 1 }=\frac { 1 }{ 5 } log(x+1)+\frac { 1 }{ 10cos(\frac { \pi }{ 5 } ) } { J }_{ 1 }-\frac { 1 }{ 10cos(\frac { 2\pi }{ 5 } ) } { J }_{ 2 }\)

\({ J }_{ 1 }=\int { \frac { (x+1)dx }{ ({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1) } } =\frac { 1 }{ 2 } (\int { \frac { 2x-2cos\frac { \pi }{ 5 } }{ { x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1 } dx } )+(1+cos\frac { \pi }{ 5 } )\int { \frac { dx }{ { ( x -cos\frac { \pi }{ 5 } ) }^{ 2 }+{ (sin\frac { \pi }{ 5 } })^{ 2 } } }\)

In the expression for \({J}_{1}\) in the first integral put \(({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)=t\) and solve and the second integral can be solved putting \((x -cos\frac { \pi }{ 5 } )=tan(t)sin(\frac{\pi}{5})\) and solve to get :

\({J}_{1}=\frac{1}{2}ln({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)+(\frac { 1+cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } ){ tan }^{ -1 }(\frac { x-cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } )\)

Similarly \({ J }_{ 2 }=\int { \frac { (x+1)dx }{ ({ x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1) } } =\frac { 1 }{ 2 } (\int { \frac { 2x+2cos\frac {2 \pi }{ 5 } }{ { x }^{ 2 }+2xcos\frac {2\pi }{ 5 } +1 } dx } )+(1-cos\frac {2\pi }{ 5 } )\int { \frac { dx }{ { ( x +cos\frac { 2\pi }{ 5 } ) }^{ 2 }+{ (sin\frac { 2\pi }{ 5 } })^{ 2 } } }\)

Solving it similarly we get :

\({J}_{2}=\frac{1}{2}ln({ x }^{ 2 }+2xcos\frac {2\pi }{ 5 } +1)+(\frac { 1-cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } ){ tan }^{ -1 }(\frac { x+cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } )\)

Putting back all the values we get I as :

\(I=\frac { -2 }{ { x }^{ 2 } } -\frac { 1 }{ 5 } log(x+1)-\frac { 1 }{ 20cos\frac { \pi }{ 5 } } ln({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)+\frac { 1 }{ 20cos\frac { 2\pi }{ 5 } } ln({ x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1)-\frac { 1+cos\frac { \pi }{ 5 } }{ 10sin\frac { \pi }{ 5 } cos\frac { \pi }{ 5 } } { tan }^{ -1 }(\frac { x-cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } )\\+\frac { 1-cos\frac { 2\pi }{ 5 } }{ 10sin\frac { 2\pi }{ 5 } cos\frac { 2\pi }{ 5 } } { tan }^{ -1 }(\frac { x+cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } )\)

Ronak Agarwal - 3 years, 3 months ago

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Awesome Solution...But can't it get shorter, bro?

Tushar Gopalka - 3 years, 3 months ago

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There's no way to get it shorter.

Ronak Agarwal - 3 years, 3 months ago

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Very much complicated bro :D Anyways nice solution

Kïñshük Sïñgh - 3 years, 3 months ago

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Such integrals are always complicated.

Ronak Agarwal - 3 years, 3 months ago

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@Ronak Agarwal Yeah u r rite

Kïñshük Sïñgh - 3 years, 3 months ago

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Yeah i agreed... Please solve this

Kïñshük Sïñgh - 3 years, 3 months ago

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@Calvin Lin i cant solve this please help

Hemant Khatri - 3 years, 3 months ago

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