Question

$\underset{0}{\overset{3}{\int }}\frac{3x+1}{x^2+9}dx=$

(a) $\frac{\mathrm{\pi }}{12}+\log \left(2\sqrt{2}\right)$

(b) $\frac{\mathrm{\pi }}{2}+\log \left(2\sqrt{2}\right)$

(c) $\frac{\mathrm{\pi }}{6}+\log \left(2\sqrt{2}\right)$

(d) $\frac{\mathrm{\pi }}{3}+\log \left(2\sqrt{2}\right)$

Answer 1

$\left(\mathrm{a}\right)\frac{\mathrm{\pi }}{12}+\log \left(2\sqrt{2}\right)$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^3\frac{3x+1}{x^2+9}dx \\ I={\int }_0^3\frac{3x}{x^2+9}dx+{\int }_0^3\frac{1}{x^2+9}dx \\ {I}_1={\int }_0^3\frac{3x}{x^2+9}dx\mathrm{and}{I}_2={\int }_0^3\frac{1}{x^2+9}dx \\ \mathrm{Putting}{x}^2+9=t\mathrm{in}{I}_1 \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \mathrm{When}x\to 0;t\to 9 \\ \mathrm{and}x\to 3;t\to 18 \\ \therefore I={\int }_9^{18}\frac{3dt}{2t}+{\int }_0^3\frac{1}{x^2+9}dx \\ =\frac{3}{2}{\int }_9^{18}\frac{dt}{t}+{\int }_0^3\frac{1}{x^2+3^2}dx \\ =\frac{3}{2}{\left[\log \left(t\right)\right]}_9^{18}+\frac{1}{3}{\left[{\tan }^{-1}\left(\frac{x}{3}\right)\right]}_0^3 \\ =\frac{3}{2}\left[\log 18-\log 9\right]+\frac{1}{3}\left(\frac{\mathrm{\pi }}{4}-0\right) \\ =\frac{3}{2}\left[\log \frac{18}{9}\right]+\frac{\mathrm{\pi }}{12} \\ =\frac{3}{2}\left[\log 2\right]+\frac{\mathrm{\pi }}{12} \\ =\log \left(\sqrt{8}\right)+\frac{\mathrm{\pi }}{12} \\ =\log \left(2\sqrt{2}\right)+\frac{\mathrm{\pi }}{12} \\ =\frac{\mathrm{\pi }}{12}+\log \left(2\sqrt{2}\right) \end{gathered}$$