#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 16 maths

$\frac{\pi}{12}+\log 2(\sqrt{2})$

Given:

$\int_{0}^{3} \frac{3 x+1}{x^{2}+9} d x$

Hint:

Using $\int \frac{d x}{x} \text { and } \int \frac{1}{1+x^{2}} d x$

Explanation:

Let

\begin{aligned} &I=\int_{0}^{3} \frac{3 x+1}{x^{2}+9} d x \\\\ &=\int_{0}^{3} \frac{3 x}{x^{2}+9} d x+\int_{0}^{3} \frac{1}{x^{2}+9} d x \end{aligned}

\begin{aligned} &= \int_{0}^{3} \frac{3 x}{x^{2}+9} d x+\int_{0}^{3} \frac{1}{x^{2}+9} d x \\\\ &=\left[\frac{3}{2} \log \left|x^{2}+9\right|+\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]_{0}^{3} \end{aligned}

\begin{aligned} &=\frac{3}{2} \log 18+\frac{1}{3} \tan ^{-1} 1-\frac{3}{2} \log 9+\frac{1}{3} \tan ^{-1} 0 \\\\ &=\frac{3}{2}(\log 18-\log 9)+\frac{1}{3} \tan ^{-1}\left(\tan \frac{\pi}{4}\right)+\frac{1}{3} \tan ^{-1}(\tan 0) \end{aligned}

\begin{aligned} &=\frac{3}{2}\left(\log \frac{18}{9}\right)+\frac{1}{3} \times \frac{\pi}{4}+\frac{1}{3} \times 0, \quad\left[\therefore \log m-\log n=\log \frac{m}{n}\right] \\\\ &=\frac{3}{2} \log 2+\frac{\pi}{12}+0 \end{aligned}

\begin{aligned} &=\log (2)^{3 / 2}+\frac{\pi}{12} \\\\ &=\frac{\pi}{12}+\log 2(\sqrt{2}) \end{aligned}

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