Calculus: Early Transcendentals 8th Edition

$$\displaystyle\int_{2}^{4} \frac{x+2}{x^{2}+3x-4}\thinspace dx = \frac{\ln{48}}{5}$$
We will begin by splitting the fraction into partial fractions: $$\frac{x+2}{x^{2}+3x-4} = \frac{x+2}{(x+4)(x-1)}=\frac{A}{x+4}+\frac{B}{x-1}$$ Solving $$\frac{(x+2)(x^{2}+3x-4)}{x^{2}+3x-4}=\frac{A(x^{2}+3x-4)}{x+4}+\frac{B(x^{2}+3x-4)}{x-1}$$ Therefore $$x+2=A(x-1)+B(x+4)$$ $$A+B=1$$ $$-A+4B=2$$ $$A=\frac{2}{5},B=\frac{3}{5} \rightarrow \frac{\frac{2}{5}}{x+4}+\frac{\frac{2}{5}}{x-1}$$ $$\displaystyle\int_{2}^{4} \frac{x+2}{x^{2}+3x-4}\thinspace dx = \displaystyle\int_{2}^{4} \bigg(\frac{2/5}{x+4}+\frac{3/5}{x-1}\bigg)\thinspace dx$$ $$=\Bigg[\frac{2}{5}\ln{|x+4|}+\frac{3}{5}\ln{|x-1|}\Bigg]^{4}_2$$ $$=\frac{2}{5}\ln{8}+\frac{3}{5}\ln{3}-\frac{2}{5}\ln{6}-\frac{3}{5}\ln{1}$$ $$=\frac{1}{5}(ln{8^{2}}+\ln{3^{3}}-\ln{6^{2}})$$ $$=\frac{1}{5}\ln{\bigg(\frac{8^{2}.3^{3}}{6^{2}}}\bigg)$$ $$=\frac{\ln{48}}{5}$$