Question

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

Answer 1

${\int }_0^3\frac{3x+1}{x^2+9}dx$

$={\int }_0^3\frac{3x}{x^2+9}dx+{\int }_0^3\frac{dx}{x^2+9}$

$=\frac{3}{2}{\int }_0^3\frac{2x}{x^2+9}dx+{\int }_0^3\frac{dx}{x^2+9}$

$=\frac{3}{2}{\left[\log (x^2+9)\right]}_0^3+\frac{1}{3}{\left[{\tan }^{-1}\frac{x}{3}\right]}_0^3$

$=\frac{3}{2}[\log (3^2+9)-\log (0+9)]+\frac{1}{3}[{\tan }^{-1}\frac{3}{3}-{\tan }^{-1}0]$

$=\frac{3}{2}[\log 18-\log 9]+\frac{1}{3}[{\tan }^{-1}1-{\tan }^{-1}0]$

$=\frac{3}{2}\left[\log \frac{18}{9}\right]+\frac{1}{3}[\frac{\pi }{4}-0]$

$=\frac{3}{2}\log 2+\frac{\pi }{12}$