# What is the antiderivative of the function y, given by y=x^2/(x^2-1)?

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### 2 Answers

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

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

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

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

==> Int y= Int dx + Int (1/(x^2-1) dx

== Int y= x + Int 1/(x^2-1) dx.............(1)

We need to find the integral of 1/(x^2-1)

First we will simplify by using partial fractions.

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

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

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

==> A-B = 1==>

==> A+B = 0 ==> 2A=1 ==> A= 1/2 ==> B= -1/2

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

==> Int 1/(x^2-1) = (1/2) l ln (x-1) - ln (x+1)

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

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

**==> Int y= x + ln [ (x-1)/(x+1)]^1/2 + c= x+ ln sqrt(x-1)/(x+1) + C**

To determine the antiderivative of the function, we'll have to calculate the indefinite integral of the given function.

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

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

We notice that the denominator x^2 - 1 is a difference of 2 squares:

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

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

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

1 = Ax + A + Bx - B

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

Comparing, we'll get:

A+B = 0

A-B = 1

-2B = 1 => B = -1/2 => A = 1/2

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

We'll integrate both sides:

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

Int dx/(x^2 - 1) = (1/2)*[ln|x-1| - ln|x+1|] + C

Int dx/(x^2 - 1) = (1/2)*[ln|x-1|/|x+1|] + C

Int dx/(x^2 - 1) = ln sqrt[|x-1|/|x+1|] + C

**The antiderivative of the given function is: Int x^2 dx/(x^2-1) = x + ln sqrt[|x-1|/|x+1|] + C**