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Math Help - Integrate

  1. #1
    Member VitaX's Avatar
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    Integrate

    \int \frac{x}{25+16x^2} dx

    u=25+16x^2

    du=32xdx

    \frac{1}{32}du=x dx

    \frac{1}{32}\int \frac{1}{u} du \rightarrow \frac{1}{32}[1] \rightarrow \frac{1}{32} + C

    Where did I go wrong or miss something?
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  2. #2
    Newbie likearollingstone's Avatar
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    from (1/32)S(1/u).du * = (1/32)ln(u)

    then sub u in?

    = (1/32)ln(25+16(x^2))

    = [ln(25+16(x^2))]/32

    Is it this?

    *the S is the integral sign
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by likearollingstone View Post
    from (1/32)S(1/u).du * = (1/32)ln(u)

    then sub u in?

    = (1/32)ln(25+16(x^2))

    = [ln(25+16(x^2))]/32

    Is it this?

    *the S is the integral sign
    I think i see my mistake \int \frac{1}{x}dx \rightarrow ln|x| + C right?
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  4. #4
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    You have to substitute back for u. So it would be

    (ln(25+16x^2))/32
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  5. #5
    Newbie likearollingstone's Avatar
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    Quote Originally Posted by VitaX View Post
    I think i see my mistake \int \frac{1}{x}dx \rightarrow ln|x| + C right?
    Yep thats right
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  6. #6
    Member VitaX's Avatar
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    Quote Originally Posted by emeryj View Post
    You have to substitute back for u. So it would be

    (ln(25+16x^2))/32
    Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value?
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  7. #7
    Newbie likearollingstone's Avatar
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    Quote Originally Posted by VitaX View Post
    Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value?
    25+16(x^2) is always greater than zero so you dont really need the absolute sign but i give you permision to put it in anyway.
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