Perform a partial fraction expansion and integrate each term individually. Where are you stuck?
2X + 2
__________________ dx
(x^2 + 1 ) (x + 1)^3
Rewriting that in one line only,
(2x+2)/[(x^2 +1)(x+1)^3] dx
Simplifying it because there is a factor (x+1) that is common to both numerator and denominator,
= 2(x+1)/[(x^2 +1)(x+1)^3] dx
= 2/[(x^2 +1)(x+1)^2] dx
You want to decompose 2/[(x^2 +1)(x+1)^2] into partial fractions.
Okay.
I hope you know the process so that I don't have to explain it. (If not, then ask me.)
2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)
Multiply both sides by [(x^2 +1)(x+1)^2],
2 = (Ax +B)(x+1)^2 +C(x^2 +1)(x+1) +D(x^2 +1) ....(2)
When x=(-1), plug that into (2),
2 = (A(-1) +B)(-1+1)^2 +C(x^2 +1)(-1+1) +D((-1)^2 +1)
2 = 0 +0 +D(2)
So,
D = 2/2 = 1 ....***
When x=0, plug that into (2), ...and plug also D=1,
2 = (A(0) +B)(0+1)^2 +C(0^2 +1)(0+1) +1(0^2 +1)
2 = (B)(1) +C(1)(1) +1(1)
2 = B +C +1
2 -1 = B +C
1 = B +C ....(3)
When x=1, plug that into (2), ...and plug also D=1,
2 = (A(1) +B)(1+1)^2 +C(1^2 +1)(1+1) +1(1^2 +1)
2 = (A+B)(2)^2 +C(2)(2) +1(2)
2 = (A+B)(4) +C(4) +2
2 -2 = 4(A+B) +4C
0 = 4(A+B) +4C
0 = (A+B) +C
0 = A +B +C ...(4)
Eq.(4) minus Eq.(3),
-1 = A ....***
When x=2, plug that into (2), ...and plug also D=1 and A = -1,
2 = (-1(2) +B)(2+1)^2 +C(2^2 +1)(2+1) +1(2^2 +1)
2 = (-2 +B)(3)^2 +C(5)(3) +1(5)
2 = (-2 +B)(9) +15C +5
2 = -18 +9B +15C +5
2 +18 -5 = 9B +15C
15 = 9B +15C
Divide both sides by 3,
5 = 3B +5C ....(5)
From (3), C = 1-B, plug that into (5),
5 = 3B +5(1-B)
5 = 3B +5 -5B
5 -5 = -2B
B = 0 ....***
Susbtitute that in (3),
C = 1 -0 = 1 ...***
So we found
A = -1
B = 0
C = 1
D = 1
Hence,
2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)
2/[(x^2 +1)(x+1)^2] = (-1x +0)/(x^2 +1) +1/(x+1) +1/(x+1)^2
2/[(x^2 +1)(x+1)^2] = -x/(x^2 +1) +1/(x+1) +1/(x+1)^2 .....******
There, it is decomposed.
------------------
So,
INT.[(2x +2)/[(x^2 +1)(x+1)^3]]dx
= INT.[2 /[x^2 +1)(x+1)^2]]dx
= INT.[-x/(x^2 +1) +1/(x+1) +1/(x+1)^2]dx
= -(1/2)ln(x^2 +1) +ln(x+1) -1/(x+1) +C ....answer.