Answer
$$\frac{4}{5} \ln 2+\frac{1}{5} \ln 3 $$
Work Step by Step
Given $$\int_{2}^{4}\frac{x+2}{x^2+3x-4}dx$$
Since
\begin{align*}
\frac{x+2}{x^2+3x-4}&=\frac{A}{x +4}+\frac{B}{x-1}\\
&= \frac{A(x-1)+B(x+4)}{(x +4)(x-1)}\\
x+2&=A(x-1)+B(x+4)
\end{align*}
At $x=1 \to B=3/5$ At $x=-4 \to B=2/5$
\begin{align*}
\int_{2}^{4} \frac{x+2}{x^{2}+3 x-4} d x &=\int_{2}^{4}\left(\frac{2 / 5}{x+4}+\frac{3 / 5}{x-1}\right) d x\\
&=\left[\frac{2}{5} \ln |x+4|+\frac{3}{5} \ln |x-1|\right]_{2}^{4} \\
&=\left(\frac{2}{5} \ln 8+\frac{3}{5} \ln 3\right)-\left(\frac{2}{5} \ln 6+0\right)\\
&=\frac{4}{5} \ln 2+\frac{1}{5} \ln 3
\end{align*}
