Question

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

Answer 1

Let $I={\int }_0^2\frac{6x+3}{x^2+4}dx={\int }_0^2\frac{6x}{x^2+4}dx+{\int }_0^2\frac{3}{x^2+4}dx$
Put $x^2+4=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}$
Lower limit, when x =0, t =0 +4 =4
upper limit, when x =2, t =4+4 =8
$\therefore I={\int }_4^8\frac{6x}{t}\frac{dt}{2x}+{\int }_0^2\frac{3}{x^2+4}dx=3{\int }_4^8\frac{1}{t}dt+3{\int }_0^2\frac{1}{x^2+2^2}dx=3[logt{]}_4^8+\frac{3}{2}{\left[ta{n}^{-1}\frac{x}{2}\right]}_0^2[\because \int \frac{dx}{a+x^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a}]=3[log\left(8\right)-log\left(4\right)+\frac{3}{2}]\left[ta{n}^{-1}\frac{2}{2}\right]=3log(\frac{8}{4})+\frac{3}{2}\times \frac{\pi }{4}[\because logb-loga=log\frac{b}{a}]=3log2+\frac{3\pi }{8}$