We are tasked with showing that:

$\int_1^3 x^2 \ln(x) \, dx = 9 \ln(3) - \frac{26}{9}.$

Solution using Integration by Parts

We will solve this using integration by parts. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du.$

Let us choose:

• $u = \ln(x)$, so $du = \frac{1}{x} \, dx$,
• $dv = x^2 \, dx$, so $v = \frac{x^3}{3}$.

Now, apply the integration by parts formula:

$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx.$

Simplify the integral:

$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 \, dx.$

Now, solve the remaining integral:

$\int x^2 \, dx = \frac{x^3}{3}.$

Thus, we have:

$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \frac{x^3}{9}.$

Evaluate the Definite Integral

Now, we evaluate this expression from $1$ to $3$:

$\left[ \frac{x^3}{3} \ln(x) - \frac{x^3}{9} \right]_1^3.$

First, evaluate at $x = 3$:

$\frac{3^3}{3} \ln(3) - \frac{3^3}{9} = 9 \ln(3) - \frac{27}{9} = 9 \ln(3) - 3.$

Now, evaluate at $x = 1$:

$\frac{1^3}{3} \ln(1) - \frac{1^3}{9} = 0 - \frac{1}{9} = -\frac{1}{9}.$

Thus, the definite integral is:

$(9 \ln(3) - 3) - \left( -\frac{1}{9} \right) = 9 \ln(3) - 3 + \frac{1}{9}.$

Simplify:

$9 \ln(3) - \frac{27}{9} + \frac{1}{9} = 9 \ln(3) - \frac{26}{9}.$

Conclusion

We have successfully shown that:

$\int_1^3 x^2 \ln(x) \, dx = 9 \ln(3) - \frac{26}{9}.$

Would you like a breakdown of any step or further clarifications?

Related Questions:

1. What are some typical applications of integration by parts in physics or engineering?
2. How would you approach a similar integral with different limits, like $\int_0^2 x^2 \ln(x) \, dx$?
3. Can integration by parts be applied multiple times, and how would that affect solving complex integrals?
4. How do you solve integrals involving logarithmic and trigonometric functions together?
5. What is the significance of the natural logarithm in calculus and its use in integrals?

Tip:

When choosing functions for integration by parts, it's helpful to differentiate logarithmic functions and integrate polynomials, as they simplify with successive operations.