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

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}

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