University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{-\ln |x+1|}{2}+\dfrac{3\ln |x+3|}{2} +c$
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ $\int \dfrac{x}{x^2+4x+3}=\int \dfrac{x}{(x+1)(x+3)}$ Write the integral into partial fractions as follows: $\int \dfrac{x}{(x+1)(x+3)}=\int [-\dfrac{1}{2(x+1)}+\dfrac{3}{2(x+3)}]dx$ Hence, $\int [-\dfrac{1}{2(x+1)}+\dfrac{3}{2(x+3)}]dx=\dfrac{-\ln |x+1|}{2}+\dfrac{3\ln |x+3|}{2} +c$