Integrate ∫ 1/(x^2 + 3x + 2) dx Step by Step

Step by step

The question

Question: find the integral of 1, all over x² + 3x + 2.

∫ 1x² + 3x + 2 dx =

1 Factor the bottom

First factor the bottom.

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

x² + 3x + 2 = (x + 2)(x + 1)

2 × 1 = 2, 2 + 1 = 3

2 Split the fraction

Two simple fractions can add up to this one: A over (x + 2), plus B over (x + 1).

We find the numbers A and B.

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

3 Clear the bottoms

Multiply both sides by (x + 2)(x + 1).

On the left the bottom goes.

On the right each fraction keeps the other bracket.

1 = A(x + 1) + 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: 1 = A(−1) + B(0)

Divide by −1: A = −1.

A = 1−1

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

x = −1: 1 = B(1) + A(0)

Divide by 1: B = 1.

B = 11

5 Integrate

Now the integral is two easy pieces.

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

= −ln|x + 2| + ln|x + 1| + C

6 Check

Let's check: add the two fractions back over one bottom, (x + 2)(x + 1).

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

Open the brackets and add: the top is 1 again.

Right!

= 1(x + 2)(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