Integrate ∫ (x + 7)/(x^2 - x - 6) dx Step by Step

Step by step

The question

Question: find the integral of (x + 7), all over x² − x − 6.

∫ x + 7x² − x − 6 dx =

1 Factor the bottom

First factor the bottom.

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

x² − x − 6 = (x + 2)(x − 3)

2 × (−3) = −6, 2 + (−3) = −1

2 Split the fraction

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

We find the numbers A and B.

x + 7(x + 2)(x − 3) = A(x + 2) + 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 ∫ x + 7x² − x − 6 dx = ln|x² − x − 6| ✗

3 Clear the bottoms

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

On the left the bottom goes.

On the right each fraction keeps the other bracket.

x + 7 = A(x − 3) + 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: 5 = A(−5) + B(0)

Divide by −5: A = −1.

A = 5−5 = −1

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

x = 3: 10 = B(5) + A(0)

Divide by 5: B = 2.

B = 105 = 2

5 Integrate

Now the integral is two easy pieces.

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

= −ln|x + 2| + 2ln|x − 3| + C

6 Check

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

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

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

Right!

= x + 7(x + 2)(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