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  • Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 22

Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 22

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Answer:  \log\left ( \frac{4}{e} \right )

Given:  \int_{0}^{1} \log (1+x) d x

Hint: Apply the formula of  \int u v d x

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

\begin{aligned} &=\int_{0}^{1} \log (1+x) \cdot 1 d x \\ & \end{aligned}

=[\log (1+x) x]_{0}^{1}-\int_{0}^{1} \frac{x}{1+x} d x \\

{\left[\because \int u v d x=u \int v d v-\int \frac{d}{d x} u \int v d x\right]}

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

=[\log (1+x) x]_{0}^{1}-\int_{0}^{1} 1-\frac{1}{1+x} d x

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

=\log 2-\log 0-[(1-\log 2)-(0-\log 1)]

We know  \log1 = 0

\begin{aligned} &\therefore \log 2-1+\log 2=2 \log 2-1 \\ & \end{aligned}

=\log 4-\log e

\begin{aligned} &\left(\begin{array}{l} \therefore \log e=1, a \log b=\log b^{a} \\ \therefore \log a-\log b=\log \frac{a}{b} \end{array}\right) \\ & \end{aligned}

=\log \left(\frac{4}{e}\right)

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