Question

Please help me do it

Answer 1

Step-by-step Step-by-step explanation:

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

`(4x)/(x^2+4x+3) = (4x)/((x+3)(x+1))`

`\:\:\:\:\:\:\:\:\:\:\:\:\:= (A)/(x+3) + (B)/(x+1)`

`\:\:\:\:\:\:\:\:\:\:\:\:\:= (Ax + A + Bx + 3B)/(x^2+4x+3)`

`\:\:\:\:\:\:\:\:\:\:\:\:\:=((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

`(4x)/(x^2+4x+3) = (6)/(x+3) - (2)/(x+1)`

We can now easily integrate this expression as follows:

`\displaystyle \int{(4x)/(x^2+4x+3)}dx = 6\int{(dx)/(x+3)} - 2\int{(dx)/(x+1)}`

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