Integrate ∫ (4x + 2)/(x^2 - 2x

Step by step

The question

Question: find the integral of (4x + 2), all over x² − 2x − 3.

∫ 4x + 2x² − 2x − 3 dx =

1 Factor the bottom

First factor the bottom.

We need two numbers that multiply to −3 and add to −2: 1 and −3.

x² − 2x − 3 = (x + 1)(x − 3)

1 × (−3) = −3, 1 + (−3) = −2

2 Split the fraction

Two simple fractions can add up to this one: A over (x + 1), plus B over (x − 3).

We find the numbers A and B.

4x + 2(x + 1)(x − 3) = A(x + 1) + B(x − 3)

Careful!

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

Here it is not, so we split.

Careful: not ∫ 4x + 2x² − 2x − 3 dx = ln|x² − 2x − 3| ✗

3 Clear the bottoms

Multiply both sides by (x + 1)(x − 3).

On the left the bottom goes.

On the right each fraction keeps the other bracket.

4x + 2 = A(x − 3) + 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: −2 = A(−4) + B(0)

Divide by −4: A = 1/2.

A = −2−4 = 12

Put in x = 3: it makes (x − 3) zero, so the A part is gone.

x = 3: 14 = B(4) + A(0)

Divide by 4: B = 7/2.

B = 144 = 72

5 Integrate

Now the integral is two easy pieces.

∫ 4x + 2(x + 1)(x − 3) dx = ∫ (12(x + 1) + 72(x − 3)) 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 1/2 × ln|x + 1| + 7/2 × ln|x − 3| + C.

= 12ln|x + 1| + 72ln|x − 3| + C

6 Check

Let's check: add the two fractions back over one bottom, (x + 1)(x − 3).

12(x + 1) + 72(x − 3) = 12(x − 3) + 72(x + 1)(x + 1)(x − 3)

Open the brackets and add: the top is 4x + 2 again.

Right!

= 4x + 2(x + 1)(x − 3)

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