Question

Please help me do it

Answer 1

Step-by-step Explanation:

We are going to solve this problem using integration by parts. We can write the integrand as

$\dfrac{4x}{x^2+4x+3} = \dfrac{4x}{(x+3)(x+1)}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:= \dfrac{A}{x+3} + \dfrac{B}{x+1}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:= \dfrac{Ax + A + Bx + 3B}{x^2+4x+3}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:=\dfrac{(A+B)x + (A+3B)}{x^2+4x+3}$

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

$\dfrac{4x}{x^2+4x+3} = \dfrac{6}{x+3} - \dfrac{2}{x+1}$

We can now easily integrate this expression as follows:

$\displaystyle \int{\dfrac{4x}{x^2+4x+3}}dx = 6\int{\dfrac{dx}{x+3}} - 2\int{\dfrac{dx}{x+1}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\:= 6\ln|x+3| - 2\ln|x+1| + C$