sinclairharry
New member
- Joined
- Mar 22, 2015
- Messages
- 17
Question is
\(\displaystyle \int \frac{2x^2-3x-12}{\left(2x-1\right)\left(x^2+3\right)}dx\)
and then i would apply the sub intergral rule so
\(\displaystyle =\int \frac{x\left(2x-3\right)-12}{2\left(x^2+3\right)u}du\)
\(\displaystyle \frac{1}{2}\int \frac{6\left(u+1\right)}{u\left(u+2\right)+13}-\frac{4}{u}du\)
and then to apply the sum rule to get
\(\displaystyle =\frac{1}{2}\left(3\ln \left(u\left(u+2\right)+13\right)-4\ln \left(u\right)\right)\)
meaning my answer was
\(\displaystyle =\frac{3\ln \left(\left(2x-1\right)\left(2x+1\right)+13\right)-4\ln \left(2x-1\right)}{2}+C\)
but it seems to be incorrect can anyone point out my error ?
cheers
\(\displaystyle \int \frac{2x^2-3x-12}{\left(2x-1\right)\left(x^2+3\right)}dx\)
and then i would apply the sub intergral rule so
\(\displaystyle =\int \frac{x\left(2x-3\right)-12}{2\left(x^2+3\right)u}du\)
\(\displaystyle \frac{1}{2}\int \frac{6\left(u+1\right)}{u\left(u+2\right)+13}-\frac{4}{u}du\)
and then to apply the sum rule to get
\(\displaystyle =\frac{1}{2}\left(3\ln \left(u\left(u+2\right)+13\right)-4\ln \left(u\right)\right)\)
meaning my answer was
\(\displaystyle =\frac{3\ln \left(\left(2x-1\right)\left(2x+1\right)+13\right)-4\ln \left(2x-1\right)}{2}+C\)
but it seems to be incorrect can anyone point out my error ?
cheers
Last edited by a moderator: