# yet another integration problem

Hi.I've been solving integrals in Piskunov's and I got to this integral .
$\int \frac {7x+1}{6x^2+x-1}\,dx$
so I evaluated the integral, but didn't seem to get the answer.
The answer the book suggests is $$\frac{2}{3} \ln {(3x-1)} + \frac {1}{2} \ln {(2x+1)} + C$$
but I've come to this thing: $$\frac{7}{12} [\ln {(3x-1)}+\ln {(2x+1)}-\frac{1}{12}\ln {\frac {x-\frac{1}{3}}{x+\frac{1}{2}}}]$$
Just help with simplification.
Thank you.

#### skipjack

Forum Staff
Can you post how you obtained the partial fractions so that any slips can be identified?

#### mathman

Forum Staff
Partial explanation: $\ln\frac{x-\frac{1}{3}}{x+\frac{1}{2}}=\ln(3x-1)-\ln(2x+1)-\ln(3)+\ln(2)$. You should check to see why the coefficients don't match the answer.

Can you post how you obtained the partial fractions so that any slips can be identified?
I found it.It's solved with partial fractions.

#### mathman

Forum Staff
$\ln(\frac{x-1/3}{x+1/2})=\ln(x-1/3)-\ln(x+1/2)=\ln(\frac{3x-1}{3})-\ln(\frac{2x+1}{2})=\ln(3x-1)-\ln(3)-\ln(2x+1)+\ln(2)$

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