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

Let's compute the integral:

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

For

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

, we use integration by parts:

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

, with u1 =

\ln (x)

and v2' =

x^{3}

Let's differentiate u1:

{(\ln (x))}^{'}

\frac{1}{x}

Let's integrate v2':

\int(x^{3})dx

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

Let's calculate

u1' \cdot v2

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

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

Let's compute the integral

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

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

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

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

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

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

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

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

▷Answer: $\frac{x^{4} \cdot (4 \cdot \ln (x) - 1)}{16} + C$ , with C being any constant