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Math Help - finding integral of...

  1. #1
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    finding integral of...

    i can't seem to figure out how to get this integral.
    the integrla of (x+1)/(x^2+2x+2)
    the answer is .5ln(x^2+2x+2)+C
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  2. #2
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    Hello, xslim12!

    i can't seem to figure out how to get this integral.

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

    The answer has \ln in it . . . didn't it make you suspicious?


    Let: u \:=\:x^2 + 2x + 2\quad\Rightarrow\quad du \:=\:(2x + 2)\,dx\quad\Rightarrow\quad (x+1)dx \:=\:\frac{1}{2}du


    Substitute: . \int\frac{\overbrace{(x+1)\,dx}^{\frac{1}{2}\,du}}  {\underbrace{x^2+2x+2}_u} \;=\;\frac{1}{2}\int\frac{du}{u} . . . . Got it?

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, xslim12!


    The answer has \ln in it . . . didn't it make you suspicious?


    Let: 2x + 2)\,dx\quad\Rightarrow\quad (x+1)dx \:=\:\frac{1}{2}du" alt="u \:=\:x^2 + 2x + 2\quad\Rightarrow\quad du \:=\2x + 2)\,dx\quad\Rightarrow\quad (x+1)dx \:=\:\frac{1}{2}du" />


    Substitute: . \int\frac{\overbrace{(x+1)\,dx}^{\frac{1}{2}\,du}}  {\underbrace{x^2+2x+2}_u} \;=\;\frac{1}{2}\int\frac{du}{u} . . . . Got it?

    omg your a life saver. Integration by substitution, i completely forgot about that!
    accidentally said by parts LOL :-p
    thanks again man
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