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Indefinite Integral: `int(x^2)/(4-3x-2x^2)^(3/2) dx`

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Sukriye3 | eNotes Newbie

Posted October 6, 2013 at 2:41 AM via web

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Indefinite Integral:

`int(x^2)/(4-3x-2x^2)^(3/2) dx`

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aruv | High School Teacher | Valedictorian

Posted October 6, 2013 at 10:19 AM (Answer #1)

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`int(x^2dx)/(4-3x-2x^2)^(3/2)`

`=2^(-3/2)int(x^2dx)/(41/16-(x+3/4)^2)^(3/2)`

`=2^(-3/2)int((sqrt(41/16)sin(t)-3/4)^2sqrt(41/16)cos(t)dt)/((41/16)^(3/2)cos^3(t))`

`=(2^(-3/2)16)/41int(41/16sin^2(t)+9/16-(6sqrt(41))/16sin(t))/(cos^2(t))dt`

`=(2^(-3/2)16)/41int((50/16)sec^2(t)-41/16-((6sqrt(41))/16)sin(t)/(cos^2(t)))dt`

`=((2^(-3/2)16)/41){(50/16)tan(t)-41/16t-((6sqrt(41))/16)/(cos(t))}+C`

`=2^(-3/2){50/41tan(t)-t-(6sqrt(41)sec(t))/41}+C`

`where`

`x+3/4=sqrt(41/16)sin(t)`

`t=sin^(-1)(sqrt(16/41)(x+3/4))`

`=sin^(-1)((4x+3)/sqrt(41))`

``

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