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

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Answer

$F (1) \approx 2.303$

Steps

Step 1 ：Let $u = x^{2} + 4 x + 5$ . Then $d u = (2 x + 4) d x$ .

Step 2 ： $F (x) = \int \frac{2 x + 4}{x^{2} + 4 x + 5} d x = \int \frac{1}{u} d u$ .

Step 3 ： $F (x) = \ln | u | + C = \ln | x^{2} + 4 x + 5 | + C$ .

Step 4 ：Using the initial condition $F (0) = \ln 5$ , we find that $C = 0$ .

Step 5 ：Thus, $F (x) = \ln | x^{2} + 4 x + 5 |$ .

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 ： $F (1) \approx 2.303$