Integrate the following wrt x : 2x /(x^2+3x+2)

Let : 2x /(x^2+3x+2) = 2x/(x+1)(x+2) = A/(x+1) + B/(x+2)

=> 2x = A(x+2) + B(x+1)

putting x= -1 in(1) , we get :

2(-1) = A(-1+2) => A = -2


putting x=-2 in(1) we get :


2(-2)=B(-2+1) =>B=4

therefore, 2x/x^2+3x+2 = -2/(x+1)+4/(x+2)

therefore, ∫[2x/x^2+3x+2]= -2∫dx/(x+1)+4∫dx/(x+2)

= -2log|x+1|+4log|x+2|+C Ans.


Edit: we can also do , -2log|x+1|+4log|x+2|+C

= 4log|x+2|-2log|x+1|+C

= log(x+2)^4-log(x+1)^2+C

= log[(x+2)^4/(x+1)^2]+C

integral (2 x)/(2+3 x+x^2) dx
Factor out constants:
= 2 integral x/(x^2+3 x+2) dx
Rewrite the integrand x/(x^2+3 x+2) as (2 x+3)/(2 (x^2+3 x+2))-3/(2 (x^2+3 x+2)):
= 2 integral ((2 x+3)/(2 (x^2+3 x+2))-3/(2 (x^2+3 x+2))) dx
Integrate the sum term by term and factor out constants:
= integral (2 x+3)/(x^2+3 x+2) dx-3 integral 1/(x^2+3 x+2) dx
For the integrand (2 x+3)/(x^2+3 x+2), substitute u = x^2+3 x+2 and du = 2 x+3 dx:
= integral 1/u du-3 integral 1/(x^2+3 x+2) dx
For the integrand 1/(x^2+3 x+2), complete the square:
= integral 1/u du-3 integral 1/((x+3/2)^2-1/4) dx
For the integrand 1/((x+3/2)^2-1/4), substitute s = x+3/2 and ds = dx:
= integral 1/u du-3 integral 1/(s^2-1/4) ds
The integral of 1/(s^2-1/4) is -2 tanh^(-1)(2 s):
= 6 tanh^(-1)(2 s)+ integral 1/u du
The integral of 1/u is log(u):
= 6 tanh^(-1)(2 s)+log(u)+constant
Substitute back for s = x+3/2:
= log(u)+6 tanh^(-1)(2 x+3)+constant
Substitute back for u = x^2+3 x+2:
= log(x^2+3 x+2)+6 tanh^(-1)(2 x+3)+constant
Which is equivalent for restricted x values to:
= 4 log(x+2)-2 log(x+1)+constant

∫ 2x / (x^2 + 3x + 2) dx
= ∫ 2x / [(x + 1) (x + 2)] dx

Let 2x / [(x+1) (x + 2)] = A/(x + 1) + B/(x + 2)
=> 2x = A(x + 2) + B(x + 1)
x = - 1 => A = - 2 and
x = - 2 => B = 4

=> Integral
= 4 ∫ dx/(x+2) - 2 ∫ dx/(x + 1)
= 4ln lx+2l - 2ln lx+1l + c
= ln[(x+2)^4/(x+1)^2] + c.