integ(4x+8)

Let's integrate:

\int (4 x + 8) d x

ⅰ

\int (4 x + 8) d x

\int 4 x d x + \int 8 d x

\int 4 x d x + \int 8 d x

ⓘ A constant factor stays in front of the integral. An antiderivative of

x^{n}

\frac{x^{n + 1}}{n + 1}

, for constant

n

with

n \neq -1

\int 4 x d x

4 \cdot \frac{x^{2}}{2}

■Differentiating

8 x

gives

8

, so

8 x

is an antiderivative of

8

\int 8 d x

8 x

4 \cdot \frac{x^{2}}{2} + 8 x

ⓘ Multiplying a fraction by a number multiplies its numerator:

4 \cdot \frac{x^{2}}{2}

\frac{4 x^{2}}{2}

■A fraction is simplified by dividing the numerator and the denominator by the same nonzero number:

\frac{4 x^{2}}{2}

2 x^{2}

2 x^{2} + 8 x

You can write the expression in factored form:

x \cdot (2 x + 8)

⇥Let's factorize in

ℝ

(2 x + 8)

⇥Factor out

2

, the greatest common divisor of the coefficients:

2 x + 8

2 \cdot (x + 4)

⇥Factor identified:

x + 4

▶Indefinite integrals are defined up to an additive constant C, so this yields:

2 x \cdot (x + 4) + C

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