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Math Help - Intergration problem

  1. #1
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    Intergration problem

    a) Intergrate (4x+5)/(x^2 +2x + 2)

    b) Intergrate x^2 inversetan x

    thanks =)
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by duskmantle View Post
    a) Intergrate (4x+5)/(x^2 +2x + 2)

    b) Intergrate x^2 inversetan x

    thanks =)
    Ad a) consider \frac{4x+5}{x^2+2x+2}=2\cdot \frac{2x+2}{x^2+2x+2}+\frac{1}{(x+1)^2+1}. Both terms on the left hand side are easy to integrate (by substitution).

    Ad b) I suggest using partial integration (integrate x^2, differentiate \arctan(x)), then do a polynomial division, finally apply a substition u := 1+x^2.
    Last edited by Failure; August 11th 2009 at 01:46 AM.
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  3. #3
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    um i got part a thanks

    but part b i ended up having to intergrate -1/6x ...how do i intergrate that? =(
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by duskmantle View Post
    um i got part a thanks

    but part b i ended up having to intergrate -1/6x ...how do i intergrate that? =(
    Integrating -\frac{1}{6x} would be easy

    \int -\frac{1}{6x}\,dx =-\frac{1}{6}\int \frac{1}{x}\,dx=-\frac{1}{6}\cdot\ln|x|+C

    However, I think you don't need that integral here, becaues I get

    \int \underset{\uparrow}{x^2}\cdot \underset{\downarrow}{\tan^{-1}(x)}\, dx=\frac{1}{3}x^3\tan^{-1}(x)-\frac{1}{3}\cdot\int x^3\frac{1}{1+x^2}\, dx

    And now you have \frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}. Integrating the first term is trivial and integrating the second term is done by substitution u:=1+x^2.
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