Integrate ∫ (x - 9)/(x^2 + 2x - 3) dx Step by Step

Step by step

The question

Question: find the integral of (x − 9), all over x² + 2x − 3.

∫ x − 9x² + 2x − 3 dx =

1 Factor the bottom

First factor the bottom.

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

x² + 2x − 3 = (x + 3)(x − 1)

3 × (−1) = −3, 3 + (−1) = 2

2 Split the fraction

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

We find the numbers A and B.

x − 9(x + 3)(x − 1) = A(x + 3) + 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 ∫ x − 9x² + 2x − 3 dx = ln|x² + 2x − 3| ✗

3 Clear the bottoms

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

On the left the bottom goes.

On the right each fraction keeps the other bracket.

x − 9 = A(x − 1) + B(x + 3)

4 Find A and B

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

x = −3: −12 = A(−4) + B(0)

Divide by −4: A = 3.

A = −12−4 = 3

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

x = 1: −8 = B(4) + A(0)

Divide by 4: B = −2.

B = −84 = −2

5 Integrate

Now the integral is two easy pieces.

∫ x − 9(x + 3)(x − 1) dx = ∫ (3(x + 3) − 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 + 3| − 2 ln|x − 1| + C.

= 3ln|x + 3| − 2ln|x − 1| + C

6 Check

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

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

Open the brackets and add: the top is x − 9 again.

Right!

= x − 9(x + 3)(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