Integrate ∫ 3/((x - 2)(x + 1)) dx Step by Step

Step by step

The question

Question: find the integral of 3, all over (x − 2)(x + 1).

∫ 3(x − 2)(x + 1) dx =

1 Split the fraction

Two simple fractions can add up to this one: A over (x − 2), plus B over (x + 1).

We find the numbers A and B.

3(x − 2)(x + 1) = A(x − 2) + 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 ∫ 3(x − 2)(x + 1) dx = ln|(x − 2)(x + 1)| ✗

2 Clear the bottoms

Multiply both sides by (x − 2)(x + 1).

On the left the bottom goes.

On the right each fraction keeps the other bracket.

3 = A(x + 1) + B(x − 2)

3 Find A and B

Put in x = 2: it makes (x − 2) zero, so the B part is gone.

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

Divide by 3: A = 1.

A = 33 = 1

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

x = −1: 3 = B(−3) + A(0)

Divide by −3: B = −1.

B = 3−3 = −1

4 Integrate

Now the integral is two easy pieces.

∫ 3(x − 2)(x + 1) dx = ∫ (1(x − 2) − 1(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 ln|x − 2| − ln|x + 1| + C.

= ln|x − 2| − ln|x + 1| + C

5 Check

Let's check: add the two fractions back over one bottom, (x − 2)(x + 1).

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

Open the brackets and add: the top is 3 again.

Right!

= 3(x − 2)(x + 1)

Let's look back.

First: split the fraction.

Then: clear the bottoms.

Then: find A and B.

Then: integrate.

Last: check.

Answer