To solve the integral

$\int_{1}^{e} t^2 \ln(t) \, dt$

using integration by parts, we apply the formula:

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

Step 1: Choose $u$ and $dv$

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

Step 2: Apply the formula

$\int t^2 \ln(t) \, dt = \left[\ln(t) \cdot \frac{t^3}{3}\right] - \int \frac{t^3}{3} \cdot \frac{1}{t} \, dt$

Simplify the second term: $\int \frac{t^3}{3} \cdot \frac{1}{t} \, dt = \frac{1}{3} \int t^2 \, dt$

Step 3: Evaluate the integrals

1. Compute the first term: $\left[\ln(t) \cdot \frac{t^3}{3}\right] \Big|_{1}^{e} = \frac{t^3}{3} \ln(t) \Big|_{1}^{e}$

2. Compute the second integral: $\frac{1}{3} \int t^2 \, dt = \frac{1}{3} \cdot \frac{t^3}{3} = \frac{t^3}{9}$ So the second term becomes: $\frac{t^3}{9} \Big|_{1}^{e}$

Step 4: Combine and evaluate

The result is: [ \int_{1}^{e} t^2 \ln(t) , dt = \left[\frac{t^3}{3} \ln(t) \right]{1}^{e} - \left[\frac{t^3}{9}\right]{1}^{e} ]

First term:

$\frac{t^3}{3} \ln(t) \Big|_{1}^{e} = \frac{e^3}{3} \ln(e) - \frac{1^3}{3} \ln(1)$ Since $\ln(e) = 1$ and $\ln(1) = 0$: $\frac{e^3}{3} \ln(e) - \frac{1^3}{3} \ln(1) = \frac{e^3}{3} - 0 = \frac{e^3}{3}$

Second term:

$\frac{t^3}{9} \Big|_{1}^{e} = \frac{e^3}{9} - \frac{1^3}{9} = \frac{e^3}{9} - \frac{1}{9}$

Final result:

Combine the two terms: $\int_{1}^{e} t^2 \ln(t) \, dt = \frac{e^3}{3} - \left(\frac{e^3}{9} - \frac{1}{9}\right)$ Simplify: $\int_{1}^{e} t^2 \ln(t) \, dt = \frac{e^3}{3} - \frac{e^3}{9} + \frac{1}{9}$

Factor and simplify further: $\int_{1}^{e} t^2 \ln(t) \, dt = \frac{3e^3}{9} - \frac{e^3}{9} + \frac{1}{9} = \frac{2e^3}{9} + \frac{1}{9}$

Final answer: $\boxed{\int_{1}^{e} t^2 \ln(t) \, dt = \frac{2e^3 + 1}{9}}$

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Would you like more details or have any questions? Here are some related topics:

1. How is integration by parts derived?
2. Can this method be used for other functions involving logarithms?
3. Why was $u = \ln(t)$ the best choice here?
4. What happens if you choose $u = t^2$ instead?
5. Applications of this integral in real-world contexts.

Tip: Integration by parts often simplifies problems when a logarithm or inverse trigonometric function is involved. Always choose $u$ as the term that simplifies when differentiated!