Integrate ∫ 5/(x^2 + x - 6) dx Step by Step

Step by step

The question

Question: find the integral of 5, all over x² + x − 6.

∫ 5x² + x − 6 dx =

1 Factor the bottom

First factor the bottom.

We need two numbers that multiply to −6 and add to 1: 3 and −2.

x² + x − 6 = (x + 3)(x − 2)

3 × (−2) = −6, 3 + (−2) = 1

2 Split the fraction

Two simple fractions can add up to this one: A over (x + 3), plus B over (x − 2).

We find the numbers A and B.

5(x + 3)(x − 2) = A(x + 3) + 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 ∫ 5x² + x − 6 dx = ln|x² + x − 6| ✗

3 Clear the bottoms

Multiply both sides by (x + 3)(x − 2).

On the left the bottom goes.

On the right each fraction keeps the other bracket.

5 = A(x − 2) + 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: 5 = A(−5) + B(0)

Divide by −5: A = −1.

A = 5−5 = −1

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

x = 2: 5 = B(5) + A(0)

Divide by 5: B = 1.

B = 55 = 1

5 Integrate

Now the integral is two easy pieces.

∫ 5(x + 3)(x − 2) dx = ∫ (−1(x + 3) + 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 + 3| + ln|x − 2| + C.

= −ln|x + 3| + ln|x − 2| + C

6 Check

Let's check: add the two fractions back over one bottom, (x + 3)(x − 2).

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

Open the brackets and add: the top is 5 again.

Right!

= 5(x + 3)(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