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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS3 Marks
Question
Evaluate the definite integral in Exercise:
$\int_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$
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Answer

$\text{Let}\ \text{I}=\int\limits_{0}^{2}\frac{6\text{x}+3}{\text{x}^{2}+4}\text{dx}$$\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 
$\text{I}=\text{F}(2)-\text{F}(0)$
$=\left\{3\log(2^{2}+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{2}{2}\bigg)\right\}-\left\{3\log(0+4)+\frac{3}{2}\tan^{-1}\bigg(\frac{0}{2}\bigg)\right\}$
$=3\log8+\frac{3}{2}\tan^{-1}1-3\log4-\frac{3}{2}\tan^{-1}0$
$=3\log8+\frac{3}{2}\bigg(\frac{\pi}{4}\bigg)-3\log4-0$
$=3\log\bigg(\frac{8}{4}\bigg)+\frac{3\pi}{8}$
$=3\log2+\frac{3\pi}{8}$

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