Answer
$\dfrac{x^2}{2}+\dfrac{3}{2} \ln |x+1|-\dfrac{9}{2} \ln |x+3|+C$
Work Step by Step
Consider the integral $\int \dfrac{x^3+4x^2}{x^2+4x+3}dx$
The given integral can be re-written as: $\int \dfrac{x^3+4x^2}{x^2+4x+3}dx=\int (x- \dfrac{3x}{x^2+4x+3})dx$
This implies that $\int (x- \dfrac{3x}{x^2+4x+3})dx=\int x dx+(\dfrac{3}{2})\int \dfrac{dx}{x+1}-(\dfrac{9}{2}) \int \dfrac{dx}{x+3}$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
or, $=\dfrac{x^2}{2}+\dfrac{3}{2} \ln |x+1|-\dfrac{9}{2} \ln |x+3|+C$
