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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 3 Maths Textbook Solution.

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Answer: \frac{-1}{3} \log |2-3 x|+\frac{2}{3} \sqrt{3 x-2}+c

Hint: To solve this equation we use \int \frac{1}{(a x+b)^{n}} d x formula

Given:  \int \frac{1}{2-3 x}+\frac{1}{\sqrt{3 x-2}} d x

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

\left\{\begin{array}{l} \int \frac{1}{a x+b} d x=\log (a x+b)+c \\ (a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)} \\ b \neq-1 \end{array}\right\}

\begin{aligned} &=\frac{\log |2-3 x|}{-3}+\frac{(3 x-2)^{\frac{1}{2}}}{3 \cdot \frac{1}{2}}+c \\ &=-\frac{\log |2-3 x|}{3}+\frac{2}{3} \sqrt{3 x+2}+c \\ &=\frac{-1}{3} \log |2-3 x|+\frac{2}{3} \sqrt{3 x-2}+c \end{aligned}



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