Question

$$\left[ -\frac{1}{2} \int \frac{2t+3}{t^2+3t+2}dt - \int \frac{3 dt}{t^2+3t+2} \right]$$

Answer 1

$\frac{5}{2} \ln|t+2| - \frac{7}{2} \ln|t+1| + C$

Answer 2

The given expression is a linear combination of two indefinite integrals.

1. The first integral $\int \frac{2t+3}{t^2+3t+2}dt$ is solved by recognizing that the numerator is the derivative of the denominator, leading to a logarithmic integral $\ln|t^2+3t+2|$.

2. The second integral $\int \frac{3 dt}{t^2+3t+2}$ is solved by factoring the denominator $t^2+3t+2 = (t+1)(t+2)$ and using partial fraction decomposition to split the integrand into $\frac{3}{t+1} - \frac{3}{t+2}$. Integrating these terms gives $3\ln|t+1| - 3\ln|t+2|$.

3. The results of the two integrals are substituted back into the original expression $-\frac{1}{2} I_1 - I_2$.

4. The logarithmic terms are combined using properties of logarithms, $\ln(ab) = \ln a + \ln b$ and $\ln(a/b) = \ln a - \ln b$, and by combining like terms of $\ln|t+1|$ and $\ln|t+2|$.

5. The arbitrary constants of integration from each integral are combined into a single arbitrary constant $C$.