Question

In Exercises \(1-10,\) find the indefinite integral. $$\int t^{2} \ln t d t$$

Short answer

The indefinite integral of the function \( t^{2} \ln t \) is \( \frac{1}{3}t^{3} \ln t -\frac{1}{9} t^{3} + C\), where \( C \) is the constant of integration.

Step-by-step solution

Define and Choose \(u\) and \(dv\)

Start by defining the functions \(u\) and \(dv\) which would simplify the integration. A good rule of thumb: choose \(u\) to be the function that simplifies when differentiated. In our case, choose \(u=\ln t\), because when differentiated, the natural logarithm becomes simpler. This leaves the rest of the integrand to be \(dv\), which is \(t^{2}\, dt\).

Find \(du\) and \(v\)

Now, find \(du\) by differentiating \(u\), and \(v\) by integrating \(dv\). So we have \( du = \frac{1}{t} \, dt \) and \( v = \int t^{2} \, dt = \frac{1}{3}t^{3}\).

Apply the Integration by Parts Formula

Now apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du \). Substituting \(u=\ln t\), \(v=\frac{1}{3}t^{3}\), \(du = \frac{1}{t} \, dt\) and simplifying gives: \( \int t^2 \ln t \, dt = \frac{1}{3}t^{3} \ln t - \int \frac{1}{3}t^{3} \cdot \frac{1}{t} \, dt \).

Simplify the Remaining Integral

The remaining integral can be simplified to \( \frac{1}{3} \int t^{2} \, dt = \frac{1}{9} t^{3} + C_1 \), where \( C_1 \) is the constant of integration.

Combine the Results

Combine the results of step 3 and step 4 to form the final result of the integral: \( \int t^2 \ln t \, dt = \frac{1}{3}t^{3} \ln t -\frac{1}{9} t^{3} + C\), where \( C = C_1 \) is the constant of integration.