Solution:
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 I1
$\Rightarrow 2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow \text{x}\text{ dx}=\frac{\text{dt}}{2}$
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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| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is: