Get Answers to all your Questions

header-bg qa

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

Answers (1)

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)

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support