MCQ
$\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}=$
  • $\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
  • B
    $\frac{\pi}{2}+\log\big(2\sqrt{2}\big)$
  • C
    $\frac{\pi}{6}+\log\big(2\sqrt{2}\big)$
  • D
    $\frac{\pi}{3}+\log\big(2\sqrt{2}\big)$

Answer

Correct option: A.
$\frac{\pi}{12}+\log\big(2\sqrt{2}\big)$
We have,
$\text{I}=\int\limits^3_0\frac{3\text{x}+1}{\text{x}^2+9}\text{ dx}$
$\text{I}=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$\text{I}_1=\int\limits^3_0\frac{3\text{x}}{\text{x}^2+9}\text{ dx}$ and $\text{I}_2=\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
Putting $\text{x}^2+9=\text{t}$ in  $l_1$     
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
when $\text{x}\rightarrow0;\text{t}\rightarrow9$
and $\text{x}\rightarrow3;\text{t}\rightarrow18$
$\therefore\text{I}=\int\limits^{18}_9\frac{3\text{ dt}}{2\text{ t}}+\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$
$=\frac{3}{2}\int\limits^{18}_9\frac{\text{dt}}{\text{t}}+\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{3}{2}\big[\log(\text{t})\big]^{18}_9+\frac{1}{3}\Big[\tan^{-1}\Big(\frac{\pi}{3}\Big)\Big]^3_0$
$=\frac{3}{2}\big[\log18-\log9\big]+\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{3}{2}\Big[\log\frac{18}{9}\Big]+\frac{\pi}{12}$
$=\frac{3}{2}\big[\log2\big]+\frac{\pi}{12}$
$=\log(\sqrt{8})+\frac{\pi}{12}$
$=\log(2\sqrt{2})+\frac{\pi}{12}$
$=\frac{\pi}{12}+\log(2\sqrt{2})$

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