Question

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

Answer 1

Let $I={\int }_0^2\frac{6x+3}{x^2+4}dx$
$\Rightarrow \int \frac{6x+3}{x^2+4}dx=3\int \frac{2x+1}{x^2+4}dx$
$=3\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx$
$=3\log (x^2+4)+\frac{3}{2}{\tan }^{-1}\frac{x}{2}=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(2\right)-F\left(0\right)$
$=\left\{3\log \right(2^2+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{2}{2}\right)\}-\left\{3\log \right(0+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{0}{2}\right)\}$
$=3\log 8+\frac{3}{2}{\tan }^{-1}1-3\log 4-\frac{3}{2}{\tan }^{-1}0$
$=3\log 8+\frac{3}{4}\left(\frac{\pi }{4}\right)-3\log 4-0$
$=3\log \left(\frac{8}{4}\right)+\frac{3\pi }{8}$
$=3\log 2+\frac{3\pi }{8}$