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  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 3 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 3 maths textbook solution.

Answers (1)

\frac{1}{6}\log\left ( \frac{35}{8} \right )

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: \int_{1}^{2}\frac{3x}{9x^2-1}\: dx

Solution: \int_{1}^{2}\frac{3x}{9x^2-1}\: dx

Putting 9x^2-1=t

\Rightarrow 18x \; dx=dt

\Rightarrow dx=\frac{dt}{18x}

When x=1  then t=9-1=8  and when x=2  then t=36-1=35

\begin{aligned} &=\int_{1}^{2} \frac{3 x}{9 x^{2}-1} d x=\int_{8}^{35} \frac{3 x}{t} \frac{d t}{18 x} \\ &=\frac{1}{6} \int_{8}^{35} \frac{1}{t} d t \\ &=\frac{1}{6}[\log |t|]_{8}^{35} \\ &=\frac{1}{6}[\log 35-\log 8] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a-\log b=\log \frac{a}{b}\right] \end{aligned}

=\frac{1}{6}\log\left ( \frac{35}{8} \right )

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