Answer
$$\dfrac{x^2}{2}+\dfrac{3\ln |x+1|}{2} -\dfrac{9}{2} \ln |x+3|+C$$
Work Step by Step
Apply the integration by parts formula such follows: $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Re-write the integral into partial fractions:
$\int \dfrac{x^3+4x^2}{x^2+4x+3}dx=\int (x- \dfrac{3x}{x^2+4x+3})dx$
Now, $$\int (x- \dfrac{3x}{x^2+4x+3})dx=\int x \space dx+(\dfrac{3}{2}) \times \int \dfrac{dx}{x+1}-(\dfrac{9}{2}) \times \int \dfrac{dx}{x+3} \\=\dfrac{x^2}{2}+\dfrac{3\ln |x+1|}{2} -\dfrac{9}{2} \ln |x+3|+C$$