Question

\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]

Answer 1

\[Let\ I = \int_1^2 \frac{x + 3}{x\left( x + 2 \right)} d x . Then, \]
\[I = \int_1^2 \left( \frac{x}{x\left( x + 2 \right)} + \frac{3}{x\left( x + 2 \right)} \right) d x\]
\[ \Rightarrow I = \int_1^2 \frac{dx}{\left( x + 2 \right)} + \int_1^2 \frac{3}{x\left( x + 2 \right)} d x\]
\[ \Rightarrow I = \left[ \log \left( x + 2 \right) \right]_1^2 + \frac{3}{2} \int_1^2 \left( \frac{1}{x} - \frac{1}{x + 2} \right) dx\]
\[ \Rightarrow I = \left[ \log \left( x + 2 \right) \right]_1^2 + \frac{3}{2} \left[ \log x - \log \left( x + 2 \right) \right]_1^2 \]
\[ \Rightarrow I = \log 4 - \log 3 + \frac{3}{2}\left[ \log 2 - \log 4 - 0 + \log 3 \right]\]
\[ \Rightarrow I = \log 4 - \log 3 + \frac{3}{2}\left[ - \log 2 + \log 3 \right]\]
\[ \Rightarrow I = 2 \log 2 - \log 3 + \frac{3}{2} \log 3 - \frac{3}{2} \log 2\]
\[ \Rightarrow I = \frac{1}{2} \log 2 + \frac{1}{2} \log 3\]
\[ \Rightarrow I = \frac{1}{2}\left( \log 2 + \log 3 \right)\]
\[ \Rightarrow I = \frac{1}{2} \log 6\]