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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- Nov 10th 2008, 12:27 PMLone21integration 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 - Nov 10th 2008, 12:38 PMMathstud28
$\displaystyle \begin{aligned}\int\log_8(2x+1)dx&=\frac{1}{\ln(8) }\int\ln(2x+1)dx\\

&\overbrace{\mapsto}^{z=2x+1}\frac{1}{2\ln(8)}\int \ln(z)dz\\

&=\int\ln(z)(z')dz\\

&=z\ln(z)-z+C~~~\text{IBP }u=\ln(z)\\

&=\frac{1}{2\ln(8)}\bigg[(2x+1)\ln(2x+1)-(2x+1)\bigg]+C\end{aligned}$