integral of x^2ln(9x) — step-by-step solution

1) Let's integrate:

\int(x^{2} \cdot \ln (9 x))dx

For

x^{2} \cdot \ln (9 x)

, we use integration by parts:

\int(u1 \cdot v2')dx = u1 \cdot v2 - \int(u1' \cdot v2)dx

, with u1 =

\ln (9 x)

and v2' =

x^{2}

2) Let's perform the differentiation u1:

\frac{d}{d x} (\ln (9 x))

Apply the differentiation rule:

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

with f =

9 x

\frac{d}{d x} (9 x) \cdot \frac{1}{9 x}

ℹ

\frac{d}{d x} (9 x) \cdot \frac{1}{9 x}

\frac{\frac{d}{d x} (9 x)}{9 x}

‖

\frac{d}{d x} (9 x)

9

\frac{9}{9 x}

\frac{1}{x}

Thus,

u1'

\frac{1}{x}

3) Let's integrate v2':

\int(x^{2})dx

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

4) Let's calculate

u1' \cdot v2

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

ℹ

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

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

‖

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

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

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

5) Let's integrate

u1 \cdot v2 - \int(u1' \cdot v2)dx

\ln (9 x) \cdot \frac{x^{3}}{3} - \int(\frac{x^{2}}{3})dx

ℹ

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

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

‖

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

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

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

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

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

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

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

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