Question

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

Answer 1

${\int }_0^2\left(\frac{6x+3}{x^2+4}\right)dx=3{\int }_0^2\left(\frac{2x}{x^2+4}\right)dx+3{\int }_0^2\left(\frac{1}{x^2+2^2}\right)dx=3\left[log\right(x^2+4){]}_0^2+3\times (\frac{1}{2}){\left[ta{n}^{-1}\right(\frac{x}{2}\left)\right]}_0^2=3(log8-log4)+(\frac{3}{2}\left)\right(ta{n}^{-1}1-ta{n}^{-1}0)=3log2+(\frac{3}{2})\times (\frac{\pi }{4})=3log2+(\frac{3\pi }{8})$