integral of -x^2ln(x) — step-by-step solution

1) Let's compute the integral:

\int (- x^{2} \cdot \ln (x)) d x

⇥For

- x^{2} \cdot \ln (x)

, we use 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^{2}

⇥1.1) Let's perform the differentiation u1:

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

⇥ⓘ The derivative of ln(x) is 1/x:

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

\frac{1}{x}

⇥

\frac{1}{x}

⇥1.2) Let's integrate v2':

\int (- x^{2}) d x

⇥ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int (- x^{2}) d x

- \frac{x^{3}}{3}

⇥

- \frac{x^{3}}{3}

⇥1.3) Let's calculate

u1' \cdot v2

\frac{1}{x} \cdot (- \frac{x^{3}}{3})

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

\frac{1}{x} \cdot (- \frac{x^{3}}{3})

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

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

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

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

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

⇥

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

⇥1.4) Let's calculate

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

\ln (x) \cdot (- \frac{x^{3}}{3}) - \int (- \frac{x^{2}}{3}) d x

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

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

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

■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

- \int (- \frac{x^{2}}{3}) d x

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

⇥

- \frac{\ln (x) \cdot x^{3}}{3} + \frac{1}{3} \cdot \frac{x^{3}}{3}

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

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

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

\frac{x^{3}}{9}

⇥

- \frac{\ln (x) \cdot x^{3}}{3} + \frac{x^{3}}{9}

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

- \frac{\ln (x) \cdot x^{3}}{3} + \frac{x^{3}}{9}

\frac{3 \cdot (- \ln (x) \cdot x^{3}) + x^{3}}{9}

⇥

\frac{3 \cdot (- \ln (x) \cdot x^{3}) + x^{3}}{9}

⇥ⓘ The minus sign is placed in front of the product:

3 \cdot (- \ln (x))

-3 \cdot \ln (x)

⇥

\frac{- 3 \cdot \ln (x) \cdot x^{3} + x^{3}}{9}

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

\frac{- 3 \cdot \ln (x) \cdot x^{3} + x^{3}}{9}

\frac{x^{3} \cdot (- 3 \cdot \ln (x) + 1)}{9}

⇥

\frac{x^{3} \cdot (- 3 \cdot \ln (x) + 1)}{9}

Finally you can finish the integration:

ⅰ

\int (- x^{2} \cdot \ln (x)) d x

\frac{x^{3} \cdot (- 3 \cdot \ln (x) + 1)}{9}

\frac{x^{3} \cdot (- 3 \cdot \ln (x) + 1)}{9}

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

\frac{x^{3} \cdot (1 - 3 \cdot \ln (x))}{9} + C