integral of (6ln(4x))/5 — step-by-step solution

1) Let's integrate:

\int \frac{6 \cdot \ln (4 x)}{5} d x

⇥Take the constant factor

\frac{6}{5}

outside the integral. For

\frac{6 \cdot \ln (4 x)}{5}

, we use integration by parts:

\int \frac{6}{5} \cdot u1 \cdot v2' d x = \frac{6}{5} \cdot (u1 \cdot v2 - \int u1' \cdot v2 d x)

, with u1 =

\ln (4 x)

and v2' =

1

⇥1.1) Let's perform the differentiation u1:

\frac{d}{d x} (\ln (4 x))

⇥Apply the differentiation rule:

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

with f =

4 x

⇥ⓘ The derivative of ln(x) is 1/x:

\frac{d}{d x} (\ln (4 x))

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

⇥

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

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

\frac{d}{d x} (4 x) \cdot \frac{1}{4 x}

\frac{\frac{d}{d x} (4 x) \cdot 1}{4 x}

\frac{\frac{d}{d x} (4 x)}{4 x}

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

\frac{d}{d x} (4 x)

4

⇥

\frac{4}{4 x}

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

\frac{4}{4 x}

\frac{1}{x}

⇥

\frac{1}{x}

⇥So,

u1'

corresponds 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 calculate

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

\frac{6}{5} \cdot (u1 \cdot v2 - \int u1' \cdot v2 d x)

\frac{6}{5} \cdot (\ln (4 x) \cdot x - \int 1 d x)

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

\frac{6}{5} \cdot (\ln (4 x) \cdot x - \int 1 d x)

\frac{6 \cdot (\ln (4 x) \cdot x - \int 1 d x)}{5}

■The derivative of

- x

-1

, so

- x

is an antiderivative of

-1

- \int 1 d x

- x

⇥

\frac{6 \cdot (\ln (4 x) \cdot x - x)}{5}

⇥ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{6 \cdot (\ln (4 x) \cdot x - x)}{5}

\frac{6 x \cdot (\ln (4 x) - 1)}{5}

⇥

\frac{6 x \cdot (\ln (4 x) - 1)}{5}

You can then resume the integration:

ⅰ

\int \frac{6 \cdot \ln (4 x)}{5} d x

\frac{6 x \cdot (\ln (4 x) - 1)}{5}

\frac{6 x \cdot (\ln (4 x) - 1)}{5}

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

\frac{6 x \cdot (\ln (4 x) - 1)}{5} + C