simple p(x)/q(x) integration question?

Question

i've been stuck on this little s**t for a bit; (x^2+1)/(x^2-1)
so I know you can split bottom into (x-1)(x+1) and you are suppose to manipulate to get a few ln's in the end, but i can't remember how to do it!

Answer 1

(x^2 + 1) / (x^2 - 1) =>

(x^2 - 1 + 2) / (x^2 - 1) =>

(x^2 - 1) / (x^2 - 1) + 2 / (x^2 - 1) =>

1 + 2 / ((x - 1) * (x + 1)) =>

1 + A / (x - 1) + B / (x + 1)



A / (x - 1) + B / (x + 1) = 2 / (x^2 - 1)

A * (x + 1) + B * (x - 1) = 0x + 2

Ax + Bx + A - B = 0x + 2



Ax + Bx = 0x

A + B = 0



A - B = 2



A + B + A - B = 0 + 2

2A = 2

A = 1



A - B = 2

1 - B = 2

-B = 1

B = -1



1 + 1 / (x - 1) - 1 / (x + 1)



Integrate



x + ln|x - 1| - ln|x + 1| + C =>

x + ln|(x - 1) / (x + 1)| + C

Answer 2

(x^2 + 1) / (x^2 - 1) = (x^2 - 1 + 1 + 1) / (x^2 - 1) = (x^2 - 1) / (x^2 - 1) + 2 / (x^2 - 1) = 1 + 2 / (x^2 - 1) = 1 + 2 / ((x + 1) * (x - 1))



Then using partial fraction decomposition on 2 / ((x + 1) * (x - 1)):



2 / ((x + 1) * (x - 1)) = A / (x + 1) + B / (x - 1)

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

=> A + B = 0 so A = -B

and 2 = B - A = B - (-B) = 2 * B

=> B = 1, A = -1



which gives:



(x^2 + 1) / (x^2 - 1) = 1 + 1 / (x - 1) - 1 / (x + 1)



Integrating gives:



Int[(x^2 + 1) / (x^2 - 1)]dx = Int[1 + 1 / (x - 1) - 1 / (x + 1)]dx

= x + Log(x - 1) - Log(x + 1)

= x + Log((x - 1) / (x + 1))

Answer 3

(x² + 1)/(x² - 1) = (x² - 1 + 2)/(x² - 1)



=> (x² - 1)/(x² - 1) + 2/(x² - 1)



=> 1 + 2/(x + 1)(x - 1)



Now, 2/(x + 1)(x - 1) = A/(x + 1) + B/(x - 1)



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



let x = 1 we get:



2 = 2B => B = 1



let x = -1 we get:



2 = -2A => A = -1



so, 2/(x + 1)(x - 1) = 1/(x - 1) - 1/(x + 1)



i.e. (x² + 1)/(x² - 1) = 1 + 1/(x - 1) - 1/(x + 1)



Then, ∫ (x² + 1)/(x² - 1) dx = ∫ 1 + 1/(x - 1) - 1/(x + 1) dx



=> x + ln(x - 1) - ln(x + 1) + C



or, x + ln[(x - 1)/(x + 1)] + C



:)>

Answer 4

(x^2+1)/(x^2-1)

x²+1= x²−1+1+1 =(x²−1)+2

=(x+1)²−2x

and x²−1 =(x+1)(x−1)

(x²+1)/(x²-1)={(x²−1)+2}/(x²-1)

=1+2/(x²-1)

Now split 2/(x²-1) into partial fractions as follows

Let 2/(x²-1) = A/(x+1) +B/(x−1)

→ 2/(x²-1) = {A(x−1) +B(x+1)}/{(x+1)(x−1)}

→2/(x²-1) = {(A+B)x+(B−A) }/{(x+1)(x−1)}

Comparing the coefficient of different powers of on both sides, give

A+B=0 →B=−A

and

(B−A) =2 i.e −A−A = 2 i.e A=−1

and B=−A=−(−1)=1

hence

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

=x−ln|(x+1)| +ln|(x−1)| +C

=x +ln{|(x−1)| /|(x+1)|}+ C

Answer 5

put it into partial fraction 1st. ie (x^2+1)(x^2-1) =[ (x^2 -1) +2]/(x^2-1) = (x^2-1)/(x^2-1) +2/(x^2-1)

ie = 1+ 2/(x2-1) now 2/(x^2-1) = 2/(x+1)(x-1) = a/(x+1) +b/(x-1) ie 2= a(x-1) +b(x+1)

when x=1, 2= 2b ie b=1 when x=-1, 2= -2a so a= -1

so (x^2+1)/(x^2-1) = 1 -1/(x+1) + 1/(x-1)

integrating gives, x - ln(x+1) +ln(x-1) + C