Find
\[
F(x)=\int \frac{2 x+4}{x^{2}+4 x+5} d x
\]
that satisfies condition $F(0)=\ln 5$. As your answer, please input $F(1)$ in decimal form with three significant digits after the decimal point.
Answer
\( \boxed{F(1) \approx 2.303} \)
Step 1 ：Let \( u = x^2 + 4x + 5 \). Then \( du = (2x + 4)dx \).
Step 2 ：\( F(x) = \int \frac{2x + 4}{x^2 + 4x + 5} dx = \int \frac{1}{u} du \).
Step 3 ：\( F(x) = \ln|u| + C = \ln|x^2 + 4x + 5| + C \).
Step 4 ：Using the initial condition \( F(0) = \ln 5 \), we find that \( C = 0 \).
Step 5 ：Thus, \( F(x) = \ln|x^2 + 4x + 5| \).
Step 6 ：Evaluating at \( x = 1 \), \( F(1) = \ln|1^2 + 4 \cdot 1 + 5| = \ln 10 \).
Step 7 ：Using a calculator to find the decimal form of \( \ln 10 \), we get \( F(1) \approx 2.303 \).
Step 8 ：\( \boxed{F(1) \approx 2.303} \)
