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  • Provide solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 31 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 31 textbook solution.

Answers (1)

Answer:-  2\log_{e}7

Hints:-  You must know the rules of integration .

Given:-  \int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x

Solution :  I=\int_{-2}^{2} \frac{3 x^{3}+2|x|+1}{x^{2}+|x|+1} d x

             \begin{aligned} &=\int_{-2}^{2} \frac{3 x^{3}}{x^{2}+|x|+1}+\int_{-2}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x\\ &\text { Odd function } \quad \text { Even function } \end{aligned}

          \begin{aligned} &=2 \int_{0}^{2} \frac{2|x|+1}{x^{2}+|x|+1} d x=2 \log _{e} 7 \\ &\frac{\pi}{2} \\ &\int_{0}^{1}\left(\tan ^{-1}(x)+\tan ^{-1}(1-x)-\tan ^{-1}(1-x)-\tan ^{-1}(x)\right) d x \end{aligned}

       I=\frac{\pi}{2}

       \begin{aligned} &I=\frac{16}{15} \sqrt{2} \\ &{\left[\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]} \\ &I=2 \sqrt{2}\left(\frac{4}{3}-\frac{4}{5}\right) \end{aligned}

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

Put x^{2}+|x|+1=t

      \begin{gathered} 2 x+1 d x=d t \\ I=2 \int_{0}^{2} \frac{d t}{t} \\ =2\left[\log _{e}|t|\right]_{0}^{2} \end{gathered}

        \begin{aligned} &=2\left[\log _{e}\left|x^{2}+\right| x|+1|\right]_{0}^{2} \\ &=2\left[\log _{e}(4+2+1)-\log (0+0+1)\right] \\ &=2\left[\log _{e} 7-\log _{e} 1\right] \\ &=2 \log _{e} 7 \end{aligned}

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