Question

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

Options

• \[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]
• \[\frac{\pi}{2} + \log\left( 2\sqrt{2} \right)\]
• \[\frac{\pi}{6} + \log\left( 2\sqrt{2} \right)\]

• \[\frac{\pi}{3} + \log\left( 2\sqrt{2} \right)\]

Answer 1

\[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]

\[\text{We have}, \]
\[I = \int_0^3 \frac{3x + 1}{x^2 + 9} d x\]
\[I = \int_0^3 \frac{3x}{x^2 + 9}dx + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ I_1 = \int_0^3 \frac{3x}{x^2 + 9}dx and I_2 = \int_0^3 \frac{1}{x^2 + 9}dx\]
\[\text{Putting} x^2 + 9 = t in I_1 \]
\[ \Rightarrow 2x\ dx = dt\]
\[ \Rightarrow x\ dx = \frac{dt}{2}\]
\[When\ x \to 0; t \to 9\]
\[and\ x \to 3; t \to 18\]
\[ \therefore I = \int_9^{18} \frac{3 dt}{2 t} + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ = \frac{3}{2} \int_9^{18} \frac{dt}{t} + \int_0^3 \frac{1}{x^2 + 3^2}dx\]
\[ = \frac{3}{2} \left[ \log\left( t \right) \right]_9^{18} + \frac{1}{3} \left[ \tan^{- 1} \left( \frac{x}{3} \right) \right]_0^3 \]
\[ = \frac{3}{2}\left[ \log18 - \log9 \right] + \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{3}{2}\left[ \log\frac{18}{9} \right] + \frac{\pi}{12}\]
\[ = \frac{3}{2}\left[ \log 2 \right] + \frac{\pi}{12}\]
\[ = \log\left( \sqrt{8} \right) + \frac{\pi}{12}\]
\[ = \log\left( 2\sqrt{2} \right) + \frac{\pi}{12}\]
\[ = \frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]