integrate x^3*ln(x) — step-by-step solution

1) Let's integrate:

\int x^{3} \cdot \ln (x) d x

⇥For

x^{3} \cdot \ln (x)

, apply integration by parts:

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

with u1 =

\ln (x)

and v2' =

x^{3}

⇥1.1) Let's perform the differentiation u1:

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

⇥ⓘ The derivative of

\ln (x)

\frac{1}{x}

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

\frac{1}{x}

⇥

\frac{1}{x}

⇥1.2) Let's compute the integral v2':

\int x^{3} d x

⇥ⓘ For constant

n

with

n \neq -1

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

is an antiderivative of

x^{n}

\int x^{3} d x

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

⇥

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

⇥1.3) Let's calculate

u1' \cdot v2

\frac{1}{x} \cdot \frac{x^{4}}{4}

⇥ⓘ The product of fractions is found by multiplying their numerators together and their denominators together:

\frac{1}{x} \cdot \frac{x^{4}}{4}

\frac{1 \cdot x^{4}}{x \cdot 4}

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

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

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

\frac{x^{3}}{4}

⇥

\frac{x^{3}}{4}

⇥1.4) Let's calculate

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

\ln (x) \cdot \frac{x^{4}}{4} - \int \frac{x^{3}}{4} d x

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

\ln (x) \cdot \frac{x^{4}}{4}

\frac{\ln (x) \cdot x^{4}}{4}

■Keep the constant factor outside the integral. For constant

n

with

n \neq -1

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

is an antiderivative of

x^{n}

- \int \frac{x^{3}}{4} d x

- \frac{1}{4} \cdot \frac{x^{4}}{4}

⇥

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

⇥ⓘ The product of fractions is found by multiplying their numerators together and their denominators together:

\frac{1}{4} \cdot \frac{x^{4}}{4}

\frac{1 \cdot x^{4}}{4 \cdot 4}

\frac{x^{4}}{16}

⇥

\frac{\ln (x) \cdot x^{4}}{4} - \frac{x^{4}}{16}

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

\frac{\ln (x) \cdot x^{4}}{4} - \frac{x^{4}}{16}

\frac{4 \cdot \ln (x) \cdot x^{4} - x^{4}}{16}

⇥

\frac{4 \cdot \ln (x) \cdot x^{4} - x^{4}}{16}

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

\frac{4 \cdot \ln (x) \cdot x^{4} - x^{4}}{16}

\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16}

⇥

\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16}

Finally you can finish the integral computation:

ⅰ

\int x^{3} \cdot \ln (x) d x

\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16}

\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16}

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

\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16} + C