INTEGRALS — Maths STD 12 Science — Question

$\int\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}=3\int\frac{2\text{x}+1}{\text{x}^{2}+4}\text{dx}$

$=3\int\frac{2\text{x}}{\text{x}^{2}+4}\text{dx}+3\int\frac{1}{\text{x}^{2}+4}\text{dx}$

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

By second fundamental theorem of calculus, we obtain 

$\text{I}=\text{F}(2)-\text{F}(0)$

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

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

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

$=3\log\bigg(\frac{8}{4}\bigg)+\frac{3\pi}{8}$

$=3\log2+\frac{3\pi}{8}$