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Math Help - Integral

  1. #1
    Senior Member pankaj's Avatar
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    Integral

    Evaluate
     <br />
\int_{0}^1\frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{(1-(1+x^2)\cot^{-1}x\}}dx<br />
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Evaluate
     <br />
\int_{0}^1 \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}}dx<br />
    man, this is one ugly integral! haha ...

    ok, we have this identity:  \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}}=1 + \frac{2x}{1+x^2} - \frac{1-2x \cot^{-1} x}{1-(1+x^2)\cot^{-1}x}. we also have \frac{d}{dx}(1 - (1+x^2) \cot^{-1}x) = 1 - 2x \cot^{-1} x.

    thus: \int \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}} \ dx=x + \ln \left|\frac{1+x^2}{1 - (1+x^2)\cot^{-1} x} \right| + C. the value of your definite integral is 1+\ln 2.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    man, this is one ugly integral! haha ...

    I am hungry for your beautiful integrals ....
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  4. #4
    Senior Member pankaj's Avatar
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    Here is my solution:

    <br />
\int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx

    =  <br />
\int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\frac{1}{1+x^2}+\frac{1}{1+x^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx<br />


    =  <br />
-\int_{0}^1\frac{\frac{-2x}{(1+x^2)^2}+\frac{1}{1+x^2}}{\frac{1}{1+x^2}-\cot^{-1}x}dx+\int_{0}^11dx<br />

    = \ln 2 +1
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  5. #5
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    Quote Originally Posted by pankaj View Post
    Here is my solution:

    <br />
\int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx

    =  <br />
\int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\frac{1}{1+x^2}+\frac{1}{1+x^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx<br />


    =  <br />
-\int_{0}^1\frac{\frac{-2x}{(1+x^2)^2}+\frac{1}{1+x^2}}{\frac{1}{1+x^2}-\cot^{-1}x}dx+\int_{0}^11dx<br />

    = \ln 2 +1
    this is better than my solution!
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