Integration With Partial Fractions?

Question

Hey am trying to integrate 2/(1-x)(1+x^2) by partial fractions. The answer should be -ln(1-x)+1/2ln(1+x^2)+1/tan(x)??
No idea where tan could possibly come in? is the answer wrong?
Thanks anyone who replies.

Answer 1

First divide the quotient into partial fractions

2/{(1 - x)(1 + x²)} = a/(1 - x) + (bx + c)/(1 + x²)
2 = a(1 + x²) + (bx + c)(1 - x) = a + ax² + bx - bx² + c - cx
2 = (a - b)x² + (b - c)x + (a + c)

0 = a - b
0 = b - c
2 = a + c

a = b
b = c
Therefore
a = c
a = b = c

2 = a + c = 2a
1 = a

a = b = c = 1

2/{(1 - x)(1 + x²)} = a/(1 - x) + (bx + c)/(1 + x²)
2/{(1 - x)(1 + x²)} = 1/(1 - x) + (1 + x)/(1 + x²)

Now we can integrate.

∫2/{(1 - x)(1 + x²)}dx = ∫{1/(1 - x) + (1 + x)/(1 + x²)}dx
= ∫{1/(1 - x)}dx + ∫{(1/(1 + x²)}dx + ∫{(x/(1 + x²)}dx

For the second integral, let
x = tan θ
dx = sec²θ dθ
θ = arctan x

= -∫{-1/(1 - x)}dx + ∫{(1/(1 + tan²θ)}(sec²θ dθ) + ½∫{(2x/(1 + x²)}dx
= -ln|1 - x| + ∫{(1/(sec²θ)}(sec²θ dθ) + ½ln(1 + x²) + C
= -ln|1 - x| + ∫dθ + ½ln(1 + x²) + C
= -ln|1 - x| + arctan(x) + ½ln(1 + x²) + C

Answer 2

I think you just misunderstood 1/tan x for Arc tan x which is the antiderivative of 1/(1+x^2) and pops up all the time when integrating rational fractions.

Answer 3

The setup: 2x+4=(A/x-2)+(B/x+4)+(C/x^2+4) Now you only get a straight forward denominator and sparkling up for A,B,C via looking the place the equations equivalent 0 and as quickly as you sparkling up for A,B,C you plug your solutions in to the 1st equation I confirmed and then combine those new equations