integrate sqrt(2x) — step-by-step solution

1) Let's compute the integral:

\int \sqrt{2 x} d x

⇥For

\sqrt{2 x}

, we use integration by parts:

\int u1 \cdot v2' d x = u1 \cdot v2 - \int u1' \cdot v2 d x

, with u1 =

\sqrt{2 x}

and v2' =

1

⇥1.1) Let's perform the differentiation u1:

\frac{d}{d x} (\sqrt{2 x})

⇥We differentiate using:

\frac{d}{d x} (\sqrt{f}) = \frac{d}{d x} (f) \cdot \frac{1}{2 \sqrt{f}}

(where f =

2 x

)

⇥ⓘ The derivative of

\sqrt{x}

\frac{1}{2 \sqrt{x}}

\frac{d}{d x} (\sqrt{2 x})

\frac{d}{d x} (2 x) \cdot \frac{1}{2 \cdot \sqrt{2 x}}

⇥

\frac{d}{d x} (2 x) \cdot \frac{1}{2 \cdot \sqrt{2 x}}

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{d}{d x} (2 x) \cdot \frac{1}{2 \cdot \sqrt{2 x}}

\frac{\frac{d}{d x} (2 x) \cdot 1}{2 \cdot \sqrt{2 x}}

\frac{\frac{d}{d x} (2 x)}{2 \cdot \sqrt{2 x}}

■The derivative of a constant multiple keeps that factor: that of

k u

k \cdot u'

\frac{d}{d x} (2 x)

2

⇥

\frac{2}{2 \cdot \sqrt{2 x}}

⇥ⓘ To simplify a fraction, divide its numerator and denominator by the same nonzero number:

\frac{2}{2 \cdot \sqrt{2 x}}

\frac{1}{\sqrt{2 x}}

⇥

\frac{1}{\sqrt{2 x}}

⇥So the value of

u1'

\frac{1}{\sqrt{2 x}}

⇥1.2) Let's compute the integral v2':

\int 1 d x

⇥ⓘ Differentiating

x

gives

1

, so

x

is an antiderivative of

1

\int 1 d x

x

⇥

x

⇥1.3) Let's evaluate

u1' \cdot v2

\frac{1}{\sqrt{2 x}} \cdot x

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{\sqrt{2 x}} \cdot x

\frac{1 \cdot x}{\sqrt{2 x}}

\frac{x}{\sqrt{2 x}}

⇥

\frac{x}{\sqrt{2 x}}

⇥ⓘ The root of a product is the product of the roots,

\sqrt{a b} = \sqrt{a} \sqrt{b}

\sqrt{2 x}

\sqrt{2} \cdot \sqrt{x}

⇥

\frac{x}{\sqrt{2} \cdot \sqrt{x}}

⇥We simplify the fraction:

\frac{x}{\sqrt{x}}

by the common factor:

\sqrt{x}

to get the new expression:

x^{1 - \frac{1}{2}}

⇥

\frac{x^{1 - \frac{1}{2}}}{\sqrt{2}}

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

1 - \frac{1}{2}

\frac{2 - 1}{2}

⇥

\frac{x^{\frac{2 - 1}{2}}}{\sqrt{2}}

⇥ⓘ The numerator and the denominator are computed before the division:

2 - 1

1

⇥

\frac{\sqrt{x}}{\sqrt{2}}

⇥By convention, we prefer numerical square roots in the numerator:

⇥ⅰ

\frac{\sqrt{x}}{\sqrt{2}}

\frac{\sqrt{2} \cdot \sqrt{x}}{2}

⇥

\frac{\sqrt{2} \cdot \sqrt{x}}{2}

⇥1.4) Let's calculate

u1 \cdot v2 - \int u1' \cdot v2 d x

\sqrt{2 x} \cdot x - \int \frac{\sqrt{2} \cdot \sqrt{x}}{2} d x

⇥Use the integration formula for

\frac{\sqrt{2} \cdot \sqrt{x}}{2}

\int x^{n} d x = \frac{x^{n + 1}}{n + 1}

⇥ⓘ Keep the constant factor outside the integral. An antiderivative of

x^{n}

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

, for constant

n

with

n \neq -1

- \int \frac{\sqrt{2} \cdot \sqrt{x}}{2} d x

- \frac{\sqrt{2}}{2} \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}

⇥

\sqrt{2 x} \cdot x - \frac{\sqrt{2}}{2} \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}

⇥ⓘ To multiply fractions, multiply the numerators together and the denominators together:

\frac{\sqrt{2}}{2} \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}

\frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{2 \cdot \frac{3}{2}}

⇥

\sqrt{2 x} \cdot x - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{2 \cdot \frac{3}{2}}

⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{3}{2}

\frac{2 \cdot 3}{2}

\frac{6}{2}

⇥

\sqrt{2 x} \cdot x - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{\frac{6}{2}}

⇥ⓘ To divide, multiply by the reciprocal of the divisor:

- \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{\frac{6}{2}}

- \sqrt{2} \cdot x^{\frac{3}{2}} \cdot \frac{2}{6}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{6}

\frac{1 \cdot 2}{3 \cdot 2}

\frac{1}{3}

⇥

\sqrt{2 x} \cdot x - \sqrt{2} \cdot x^{\frac{3}{2}} \cdot \frac{1}{3}

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\sqrt{2} \cdot x^{\frac{3}{2}} \cdot \frac{1}{3}

\frac{\sqrt{2} \cdot x^{\frac{3}{2}} \cdot 1}{3}

\frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{3}

⇥

\sqrt{2 x} \cdot x - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{3}

Then you can continue the integration:

ⅰ

\int \sqrt{2 x} d x

\sqrt{2 x} \cdot x - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{3}

\sqrt{2 x} \cdot x - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{3}

▶An indefinite integral is defined up to an additive constant C; we therefore obtain:

x \cdot \sqrt{2 x} - \frac{\sqrt{2} \cdot x^{\frac{3}{2}}}{3} + C