Integrate ∫ (7x - 2)/(x^2 - x - 2) dx Step by Step

Step by step

The question

Question: find the integral of (7x − 2), all over x² − x − 2.

∫ 7x − 2x² − x − 2 dx =

1 Factor the bottom

First factor the bottom.

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

x² − x − 2 = (x + 1)(x − 2)

1 × (−2) = −2, 1 + (−2) = −1

2 Split the fraction

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

We find the numbers A and B.

7x − 2(x + 1)(x − 2) = A(x + 1) + B(x − 2)

Careful!

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

Here it is not, so we split.

Careful: not ∫ 7x − 2x² − x − 2 dx = ln|x² − x − 2| ✗

3 Clear the bottoms

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

On the left the bottom goes.

On the right each fraction keeps the other bracket.

7x − 2 = A(x − 2) + 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: −9 = A(−3) + B(0)

Divide by −3: A = 3.

A = −9−3 = 3

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

x = 2: 12 = B(3) + A(0)

Divide by 3: B = 4.

B = 123 = 4

5 Integrate

Now the integral is two easy pieces.

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

= 3ln|x + 1| + 4ln|x − 2| + C

6 Check

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

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

Open the brackets and add: the top is 7x − 2 again.

Right!

= 7x − 2(x + 1)(x − 2)

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