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

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

Answers (1)

best_answer

Answer:

\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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infoexpert26

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