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Math Help - integration help

  1. #1
    Newbie
    Joined
    Nov 2008
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    integration help

    i cant seem to figure this out
    find the integral of
    log(base 8) (2x+1) dx

    i'm not sure how to approach this problem
    i assume theres some integration by parts
    so would it be like this
    intg( LOG(b8) (2x+1)dx=(x^2+x)(LOGb8) - intg(x^2+x/8)
    please help
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  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
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    Pennsylvania
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    Quote Originally Posted by Lone21 View Post
    i cant seem to figure this out
    find the integral of
    log(base 8) (2x+1) dx

    i'm not sure how to approach this problem
    i assume theres some integration by parts
    so would it be like this
    intg( LOG(b8) (2x+1)dx=(x^2+x)(LOGb8) - intg(x^2+x/8)
    please help
    \begin{aligned}\int\log_8(2x+1)dx&=\frac{1}{\ln(8)  }\int\ln(2x+1)dx\\<br />
&\overbrace{\mapsto}^{z=2x+1}\frac{1}{2\ln(8)}\int  \ln(z)dz\\<br />
&=\int\ln(z)(z')dz\\<br />
&=z\ln(z)-z+C~~~\text{IBP }u=\ln(z)\\<br />
&=\frac{1}{2\ln(8)}\bigg[(2x+1)\ln(2x+1)-(2x+1)\bigg]+C\end{aligned}
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