integral of ln(10x) — step-by-step solution

1) Let's compute the integral:

\int \ln (10 x) d x

⇥For

\ln (10 x)

, we use integration by parts:

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

, with u1 =

\ln (10 x)

and v2' =

1

⇥1.1) Let's differentiate u1:

\frac{d}{d x} (\ln (10 x))

⇥Apply the differentiation rule:

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

with f =

10 x

⇥ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (10 x))

\frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

⇥

\frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

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

\frac{d}{d x} (10 x) \cdot \frac{1}{10 x}

\frac{\frac{d}{d x} (10 x) \cdot 1}{10 x}

\frac{\frac{d}{d x} (10 x)}{10 x}

■A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (10 x)

10

⇥

\frac{10}{10 x}

⇥ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{10}{10 x}

\frac{1}{x}

⇥

\frac{1}{x}

⇥So

u1'

is equal to

\frac{1}{x}

⇥1.2) Let's compute the integral v2':

\int 1 d x

⇥ⓘ The derivative of

x

1

, so

x

is an antiderivative of

1

\int 1 d x

x

⇥

x

⇥1.3) Let's evaluate

u1' \cdot v2

\frac{1}{x} \cdot x

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

\frac{1}{x} \cdot x

\frac{1 \cdot x}{x}

\frac{x}{x}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor

x

\frac{x}{x}

1

⇥

1

⇥1.4) Let's calculate

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

\ln (10 x) \cdot x - \int 1 d x

⇥ⓘ The derivative of

- x

-1

, so

- x

is an antiderivative of

-1

- \int 1 d x

- x

⇥

\ln (10 x) \cdot x - x

Then you can continue the integration:

ⅰ

\int \ln (10 x) d x

\ln (10 x) \cdot x - x

\ln (10 x) \cdot x - x

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

x \cdot \ln (10 x) - x + C