Answer
$\dfrac{-1}{2}\ln |x+1|+\dfrac{3}{2}\ln |x+3|+c$
Work Step by Step
The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Now, we have
$I=\int \dfrac{x}{x^2+4x+3}=\int \dfrac{x}{(x+1)(x+3)}$
Need to write the integral into partial fractions.
$I=\int \dfrac{x}{(x+1)(x+3)}=\int -\dfrac{1}{2(x+1)} dx +\int \dfrac{3}{2(x+3)} dx$
Thus,
$I=\int [-\dfrac{1}{2(x+1)}+\dfrac{3}{2(x+3)}]dx$
$\implies I=\dfrac{-1}{2}\ln |x+1|+\dfrac{3}{2}\ln |x+3|+c$
