Integral of (x + 2) Cubed

We are asked to evaluate the following integral:

$$ \int (x+2)^3 \ dx $$

This can be approached in more than one way.

Method 1

We begin by introducing a substitution: u = x + 2.

$$ \int u^3 \ dx $$

The integral now has a standard elementary form:

$$ \int u^3 \ dx = \frac{u^{3+1}}{3+1} + c $$

$$ \int u^3 \ dx = \frac{u^4}{4} + c $$

Substituting back $ u = x + 2 $, we obtain:

$$ \int (x+2)^3 \ dx = \frac{(x+2)^4}{4} + c $$

Thus, the antiderivative is:

$$ F(x) = \frac{(x+2)^4}{4} + c $$

Verification: To confirm the result, we differentiate $ F(x) = \frac{(x+2)^4}{4} + c $: $$ D \left[ \frac{(x+2)^4}{4} \right] = \frac{1}{4} \cdot D[(x+2)^4] = \frac{1}{4} \cdot 4(x+2)^3 = (x+2)^3 $$ as expected.

Method 2

The integral

$$ \int (x+2)^3 \ dx $$

is a classic example of the form f ′(x) · [f(x)]ⁿ, where $ f(x) = x + 2 $, $ f'(x) = 1 $, and $ n = 3 $.

For integrals of this type, we apply the general formula:

$$ \int f'(x) \cdot [f(x)]^n \ dx = \frac{[f(x)]^{n+1}}{n+1} + c $$

Using this integration technique gives:

$$ \int (x+2)^3 \ dx = \frac{(x+2)^{3+1}}{3+1} + c $$

$$ \int (x+2)^3 \ dx = \frac{(x+2)^4}{4} + c $$

So once again, the antiderivative is:

$$ F(x) = \frac{(x+2)^4}{4} + c $$

This confirms the result obtained by the first method.

And so on.