Step-by-step Explanation:
We are going to solve this problem using integration by parts. We can write the integrand as
[tex]\dfrac{4x}{x^2+4x+3} = \dfrac{4x}{(x+3)(x+1)}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:= \dfrac{A}{x+3} + \dfrac{B}{x+1}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:= \dfrac{Ax + A + Bx + 3B}{x^2+4x+3}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:=\dfrac{(A+B)x + (A+3B)}{x^2+4x+3}[/tex]
Comparing the above term to the left hand side, we can see that
(A + B)x = 4x
A + 3B = 0
Solving for A and B, we find that A = 6 and B = -2 so our integrand becomes
[tex]\dfrac{4x}{x^2+4x+3} = \dfrac{6}{x+3} - \dfrac{2}{x+1}[/tex]
We can now easily integrate this expression as follows:
[tex]\displaystyle \int{\dfrac{4x}{x^2+4x+3}}dx = 6\int{\dfrac{dx}{x+3}} - 2\int{\dfrac{dx}{x+1}}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:= 6\ln|x+3| - 2\ln|x+1| + C[/tex]
