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Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 9

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Answer:  2\log2 -1

Given:  \int_{0}^{1} \frac{1-x}{1+x} d x

Hint: Apply integration by parts method


         \int_{0}^{1} \frac{1-x}{1+x} d x

\begin{aligned} &=\int_{0}^{1} \frac{1-x-1+1}{1+x} d x=\int_{0}^{1} \frac{2-(x+1)}{1+x} d x \\ & \end{aligned}

=\int_{0}^{1} \frac{2}{1+x} d x-\int_{0}^{1} d x

\begin{aligned} &=2[\log (1+x)]_{0}^{1}-(x)_{0}^{1} \\ & \end{aligned}

=2(\log 2-\log 1)-1 \\

=2 \log 2-1

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