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Math Help - Integration Problem

  1. #1
    Junior Member
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    Integration Problem

    Hi

    I am stuck on this problem integrating the following.

    \int\frac{arctan (x)}{1+x^2}dx

    In order to integrate should I be integrating by parts and seperating the function into arctan(x)\cdot\frac{1}{1+x^2}

    Or should I be using the substitution method to do this?

    I am very stuck - I know that \int arctan(x) = x\cdot arctan (x) - \frac {1}{2}ln(1+x^2) + c but don't know if it's relevant to help me solve??

    Thanks
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Ian1779 View Post
    Hi

    I am stuck on this problem integrating the following.

    \int\frac{arctan (x)}{1+x^2}dx

    In order to integrate should I be integrating by parts and seperating the function into arctan(x)\cdot\frac{1}{1+x^2}

    Or should I be using the substitution method to do this?

    I am very stuck - I know that \int arctan(x) = x\cdot arctan (x) - \frac {1}{2}ln(1+x^2) + c but don't know if it's relevant to help me solve??

    Thanks
    Just substitution \arctan x = u

    \int {\frac{{\arctan x}}{{1 + {x^2}}}dx}  = \left\{ \begin{gathered} \arctan x = u, \hfill \\\frac{{dx}}{{1 + {x^2}}} = du \hfill \\ <br />
\end{gathered}  \right\} = \int {udu}  = \frac{{{u^2}}}{2} + C = \frac{{{{\arctan }^2}x}}{2} + C.
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  3. #3
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    Aug 2009
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    Use integration by parts

    u=1+x^2
    du=arctan(x)
    v=1+x^2
    dv=arctan(x)

    I=integral(u*dv) -->what you're looking for

    I=uv-integral(v*du)=(1+x^2)^2-integral(v*du)

    But integral(v*du)=integral(u*dv) they're the same integral

    so,

    I=(1+x^2)^2 -I
    2*I=(1+x^2)^2

    I=((1+x^2)^2)/2 --->this is the answer
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  4. #4
    Junior Member
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    Thanks - I never thought to substitute in for arctan (x)

    Make sense now!!
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