To find the indefinite integral of \( \int t \ln (t+3) \, dt \), we can use integration by parts. Integration by parts is based on the formula:
\[
\int u \, dv = uv - \int v \, du
\]
We will choose \( u = \ln(t+3) \) and \( dv = t \, dt \). Then, we will find \( du \) and \( v \) and apply the integration by parts formula.
We need to find the indefinite integral:
\[
\int t \ln(t+3) \, dt
\]
Using the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
we choose:
\[
u = \ln(t+3) \quad \text{and} \quad dv = t \, dt
\]
First, compute \( du \):
\[
u = \ln(t+3) \implies du = \frac{1}{t+3} \, dt
\]
Next, compute \( v \):
\[
dv = t \, dt \implies v = \frac{t^2}{2}
\]
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:
\[
\int t \ln(t+3) \, dt = \frac{t^2}{2} \ln(t+3) - \int \frac{t^2}{2} \cdot \frac{1}{t+3} \, dt
\]
Simplify the remaining integral:
\[
\int \frac{t^2}{2(t+3)} \, dt
\]
This can be further simplified and integrated to:
\[
\int \frac{t^2}{2(t+3)} \, dt = \frac{1}{2} \left( \int t - \frac{9}{t+3} \, dt \right)
\]
Integrate each term separately:
\[
\int t \, dt = \frac{t^2}{2}
\]
\[
\int \frac{9}{t+3} \, dt = 9 \ln|t+3|
\]
Combine all parts to get the final integral:
\[
\int t \ln(t+3) \, dt = \frac{t^2}{2} \ln(t+3) - \frac{1}{2} \left( \frac{t^2}{2} - 9 \ln|t+3| \right) + C
\]
Simplify the final expression:
\[
\int t \ln(t+3) \, dt = \frac{t^2}{2} \ln(t+3) - \frac{t^2}{4} + \frac{9}{2} \ln|t+3| + C
\]
\[
\boxed{\int t \ln(t+3) \, dt = \frac{t^2}{2} \ln(t+3) - \frac{t^2}{4} + \frac{9}{2} \ln|t+3| + C}
\]
Answered
