Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 26 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 26 Maths Textbook Solution.

Answers (1)

best_answer

Answer:  \frac{1}{2}\log 6

Given:  \int_{1}^{2} \frac{(x+3)}{x(x+2)} d x

Hint: Do integration by parts

Solution:  \int_{1}^{2} \frac{(x+3)}{x(x+2)} d x

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

=\int_{1}^{2} \frac{1}{x} d x+\int_{1}^{2}\left(\frac{1}{x(x+2)}\right) d x \\

=\int_{1}^{2} \frac{1}{x} d x+\frac{1}{2} \int_{1}^{2}\left(\frac{x+2-x}{x(x+2)}\right) d x

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

=\frac{3}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} d x

\begin{aligned} &=\frac{3}{2}(\log x)_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ & \end{aligned}

=\frac{3}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \\

=\frac{3}{2} \log 2-\log 2+\frac{1}{2} \log 3

\begin{aligned} &=\frac{1}{2} \log 2+\frac{1}{2} \log 3=\frac{1}{2}(\log 2+\log 3) \\ & \end{aligned}

=\frac{1}{2} \log 6(\therefore \log a+\log b=\log a b)

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Latest Question