Question

Integrals

Answer 1

Have done this problem a couple times and have gotten the same answer each time. Would someone be able to check it for me?

Answer 2

http://www.wolframalpha.com/input/?i=%5Cint+%2840x-40%29%2F%283x%5E2-8x-3%29+dx

Answer 3

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[\int\frac{40x-40}{3x^2-8x-3}~dx\]
Factor the denominator, pull out a constant:
\[40\int\frac{x-1}{(3x+1)(x-3)}~dx\]
Partial fraction decomposition:
\[\begin{align*}\frac{x-1}{(3x+1)(x-3)}&=\frac{A}{3x+1}+\frac{B}{x-3}\\
x-1&=A(x-3)+B(3x+1)\\
x-1&=(A+3B)x-3A+B
\end{align*}\]
Match up the coefficients:
\[\begin{cases}A+3B=1\\
-3A+B=-1\end{cases}~~\Rightarrow~~\color{red}{A=\frac{2}{5},~B=\frac{1}{5}}\]
\[\begin{align*}\int\frac{40x-40}{3x^2-8x-3}~dx&=40\left(\frac{2}{5}\int\frac{dx}{3x+1}+\frac{1}{5}\int\frac{dx}{x-3}\right)\\
&=\frac{16}{3}\ln|3x+1|+8\ln|x-3|+C
\end{align*}\]
\(\color{blue}{\text{End of Quote}}\)