Answer
$x^2-3x+\dfrac{2}{3} \ln |x+4|+\dfrac{1}{3} \ln |x-2|+C$
Work Step by Step
Consider the integral $\int \dfrac{2x^3+x^2-21x+24}{x^2+2x-8}dx$
The given integral can be re-written as: $\int \dfrac{2x^3+x^2-21x+24}{x^2+2x-8}dx=\int (2x-3)+ \dfrac{x}{x^2+2x-8}dx$
This implies that $\int [(2x-3)+ \dfrac{x}{x^2+2x-8}]dx=\int (2x-3) dx+(\dfrac{1}{3})\int \dfrac{dx}{x-2}+(\dfrac{2}{3}) \int \dfrac{dx}{x+4}$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
or, $=x^2-3x+\dfrac{2}{3} \ln |x+4|+\dfrac{1}{3} \ln |x-2|+C$
