yet another integration problem

Nov 2018
24
1
Iran
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
Dec 2006
21,478
2,470
Can you post how you obtained the partial fractions so that any slips can be identified?
 

mathman

Forum Staff
May 2007
6,932
774
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.
 

mathman

Forum Staff
May 2007
6,932
774
$\ln(\frac{x-1/3}{x+1/2})=\ln(x-1/3)-\ln(x+1/2)=\ln(\frac{3x-1}{3})-\ln(\frac{2x+1}{2})=\ln(3x-1)-\ln(3)-\ln(2x+1)+\ln(2)$
 
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