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Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 9 maths textbook solution.

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Answer : \frac{\pi}{8} \log 2

Hints:-  You must know the integration rules of trigonometric functions and its limits

Given:-  \int_{0}^{1} \frac{\log x(1+x)}{1+x^{2}} d x

Solution : \int_{0}^{1} \frac{\log x(1+x)}{1+x^{2}} d x

             \begin{aligned} &\text { Put } x=\tan \theta \\ &d x=\sec ^{2} \theta d \theta \end{aligned}

Lower limit, x=0 \text {, then } \theta=\tan ^{-1} 0=0

Upper limit, x=1, \text { then } \theta=\tan ^{-1}=\frac{\pi}{4}

           \begin{aligned} &\therefore I=\int_{0}^{\frac{\pi}{4}} \frac{\log |1+\tan \theta|}{1+\tan ^{2} \theta} \sec ^{2} \theta d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \frac{\log |1+\tan \theta|}{\sec ^{2} \theta} \sec ^{2} \theta d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log |1+\tan \theta| d \theta \end{aligned}

          \begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \log \left|1+\tan \left(\frac{\pi}{4}-\theta\right)\right| \cdot d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log \left|1+\frac{(1-\tan \theta)}{1+\tan \theta}\right| \cdot d \theta \\ &I=\int_{0}^{\frac{\pi}{4}} \log \left|\frac{2}{1+\tan \theta}\right| \cdot d \theta \end{aligned}

        \begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \log 2 d \theta-\int_{0}^{\frac{\pi}{4}} \log |1+\tan \theta| \cdot d \theta \\ &I=\log 2[\theta]_{0}^{\frac{\pi}{4}}-I \\ &2 I=\frac{\pi}{4} \log 2 \\ &I=\frac{\pi}{8} \log 2 \end{aligned}

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