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
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
$\displaystyle \begin{aligned}\int\log_8(2x+1)dx&=\frac{1}{\ln(8) }\int\ln(2x+1)dx\\
&\overbrace{\mapsto}^{z=2x+1}\frac{1}{2\ln(8)}\int \ln(z)dz\\
&=\int\ln(z)(z')dz\\
&=z\ln(z)-z+C~~~\text{IBP }u=\ln(z)\\
&=\frac{1}{2\ln(8)}\bigg[(2x+1)\ln(2x+1)-(2x+1)\bigg]+C\end{aligned}$