Partial Fractions for \int \frac{2x+1}{4x^2+12x-7}dx

[nameVoid]

\int \frac{2x+1}{4x^2+12x-7}dx
\frac{1}{4} \int \frac{2x+1}{x^2+3x-\frac{7}{4}}dx
\frac{1}{4} \int \frac{2x+1}{(x+\frac{3}{2})^2-4}dx
u=x+\frac{3}{2}
\frac{1}{2} \int \frac{u-1}{u^2-4}du
\frac{1}{2} \int \frac{u}{u^2-4}du -\frac{1}{2}\int \frac{du}{u^2-4}
\frac{1}{4} ln|u^2-4|-\frac{1}{2}\int \frac{A}{u+2} +\frac{B}{u-2} du
-\frac{1}{2}=A(u-2)+B(u+2)
A=\frac{1}{8}
B=-\frac{1}{8}
\frac{1}{4} ln|u^2-4|+\frac{1}{8}ln|\frac{u+2}{u-2}|+C
\frac{1}{4} ln|x^2+3x-\frac{7}{4}|-\frac{1}{8} ln|\frac{x+\frac{7}{2}}{x-\frac{1}{2}}|+C

 

[gabbagabbahey]

What you've got is correct, but it can still be simplified further; \ln|u^2-4|=\ln|u-2|+\ln|u-2| and \ln\left|\frac{u+2}{u-2}\right|=\ln|u-2|-\ln|u-2|.

So,

\frac{1}{4}\ln|u^2-4|+\frac{1}{8}ln\left|\frac{u+2}{u-2}\right|=\frac{3}{8}\ln|u+2|+\frac{1}{8}\ln|u-2|