Question

${\int }_0^2\frac{6x+3}{x^2+4}$ dx

Answer 1

$I={\int }_0^2\frac{6x+3}{x^2+4}dx=3{\int }_0^2\frac{2x}{x^2+4}+3{\int }_0^2\frac{dx}{x^2+2^2}$

$I_1$ $I_2$

$\Rightarrow I_2=3{\int }_0^2\frac{dx}{x^2+2^2}={\left[\frac{3}{2}{\tan }^{-1}\frac{x}{2}\right]}_0^2$

$=\frac{3}{2}[{\tan }^{-1}\frac{2}{2}-{\tan }^{-1}\frac{0}{2}]$

$=\frac{3}{2}[{\tan }^{-1}1-0]$

$=\frac{3}{2}\times \frac{\pi }{4}$

$=\frac{3\pi }{8}$

$\Rightarrow I_1=3{\int }_0^2\frac{2x}{x^2+4}dx$

$={[3l\eta x^2+4]}_0^2$

$=3[l\eta 8+l\eta 4]$

$=3l\eta \frac{8}{4}$

$=3l\eta 2$

$\therefore I=I_1+I_2$

$=3l\eta 2+\frac{3\pi }{8}$

Hence, solved.