Integrate ∫ 4/(x^2 - 4) dx Step by Step

Step by step

The question

Question: find the integral of 4, all over x² − 4.

∫ 4x² − 4 dx =

1 Factor the bottom

First factor the bottom.

x² − 4 is a difference of two squares: x² minus 2 squared.

So it is (x − 2)(x + 2).

x² − 4 = (x + 2)(x − 2)

a² − b² = (a − b)(a + b), b = 2

2 Split the fraction

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

We find the numbers A and B.

4(x + 2)(x − 2) = A(x + 2) + 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 ∫ 4x² − 4 dx = ln|x² − 4| ✗

3 Clear the bottoms

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

On the left the bottom goes.

On the right each fraction keeps the other bracket.

4 = A(x − 2) + B(x + 2)

4 Find A and B

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

x = −2: 4 = A(−4) + B(0)

Divide by −4: A = −1.

A = 4−4 = −1

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

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

Divide by 4: B = 1.

B = 44 = 1

5 Integrate

Now the integral is two easy pieces.

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

= −ln|x + 2| + ln|x − 2| + C

6 Check

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

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

Open the brackets and add: the top is 4 again.

Right!

= 4(x + 2)(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