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  • 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)

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)

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