Integrate ∫ (5x - 1)/(x^2 - 1) dx Step by Step

Step by step

The question

Question: find the integral of (5x − 1), all over x² − 1.

∫ 5x − 1x² − 1 dx =

1 Factor the bottom

First factor the bottom.

x² − 1 is a difference of two squares: x² minus 1 squared.

So it is (x − 1)(x + 1).

x² − 1 = (x + 1)(x − 1)

a² − b² = (a − b)(a + b), b = 1

2 Split the fraction

Two simple fractions can add up to this one: A over (x + 1), plus B over (x − 1).

We find the numbers A and B.

5x − 1(x + 1)(x − 1) = A(x + 1) + B(x − 1)

Careful!

ln of the bottom works only when the top is the bottom's derivative.

Here it is not, so we split.

Careful: not ∫ 5x − 1x² − 1 dx = ln|x² − 1| ✗

3 Clear the bottoms

Multiply both sides by (x + 1)(x − 1).

On the left the bottom goes.

On the right each fraction keeps the other bracket.

5x − 1 = A(x − 1) + B(x + 1)

4 Find A and B

Put in x = −1: it makes (x + 1) zero, so the B part is gone.

x = −1: −6 = A(−2) + B(0)

Divide by −2: A = 3.

A = −6−2 = 3

Put in x = 1: it makes (x − 1) zero, so the A part is gone.

x = 1: 4 = B(2) + A(0)

Divide by 2: B = 2.

B = 42 = 2

5 Integrate

Now the integral is two easy pieces.

∫ 5x − 1(x + 1)(x − 1) dx = ∫ (3(x + 1) + 2(x − 1)) dx

One over x minus a number integrates to ln of it, like 1 over x.

∫ 1x − p dx = ln|x − p| + C

So the answer is 3 ln|x + 1| + 2 ln|x − 1| + C.

= 3ln|x + 1| + 2ln|x − 1| + C

6 Check

Let's check: add the two fractions back over one bottom, (x + 1)(x − 1).

3(x + 1) + 2(x − 1) = 3(x − 1) + 2(x + 1)(x + 1)(x − 1)

Open the brackets and add: the top is 5x − 1 again.

Right!

= 5x − 1(x + 1)(x − 1)

Let's look back.

First: factor the bottom.

Then: split the fraction.

Then: clear the bottoms.

Then: find A and B.

Then: integrate.

Last: check.

Answer