Question

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

Answer 1

Let I=$∫_0^2\frac{6x+3}{x^2+4} dx$

∫_0^2\frac{6x+3}{x^2+4} dx=3$$∫\frac{2x+1}{x^2+4}dx

∫\frac{2x+1}{x^2+4}dx+3$$∫\frac{1}{x^2+4}dx

$=3log(x^2+4)+\frac{3}{2}tan^{-1}\frac{x}{2}=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(2)-F(0)$

$={3log(2^2+4)+\frac{3}{2}tan^{-1}(\frac{2}{2})}-{3log(0+4)+\frac{3}{2}tan^{-1}(\frac{0}{2}})$

$=3log8+\frac{3}{2}tan^{-1}-3log4-\frac{3}{2}tan^{-1} 0$

$=3log8+\frac{3}{2}(\frac{π}{4})-3log4-0$

$=3log(\frac{8}{4})+\frac{3π}{8}$

$=3log2+\frac{3π}{8}$