Integrate ∫ 6/(x^2 - 9) dx Step by Step

Step by step

The question

Question: find the integral of 6, all over x² − 9.

∫ 6x² − 9 dx =

1 Factor the bottom

First factor the bottom.

x² − 9 is a difference of two squares: x² minus 3 squared.

So it is (x − 3)(x + 3).

x² − 9 = (x + 3)(x − 3)

a² − b² = (a − b)(a + b), b = 3

2 Split the fraction

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

We find the numbers A and B.

6(x + 3)(x − 3) = A(x + 3) + 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 ∫ 6x² − 9 dx = ln|x² − 9| ✗

3 Clear the bottoms

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

On the left the bottom goes.

On the right each fraction keeps the other bracket.

6 = A(x − 3) + 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: 6 = A(−6) + B(0)

Divide by −6: A = −1.

A = 6−6 = −1

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

x = 3: 6 = B(6) + A(0)

Divide by 6: B = 1.

B = 66 = 1

5 Integrate

Now the integral is two easy pieces.

∫ 6(x + 3)(x − 3) dx = ∫ (−1(x + 3) + 1(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 −ln|x + 3| + ln|x − 3| + C.

= −ln|x + 3| + ln|x − 3| + C

6 Check

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

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

Open the brackets and add: the top is 6 again.

Right!

= 6(x + 3)(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