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Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.5 Question 8 Maths Textbook Solution.

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Answer:  \frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{2}{9}(1+3 x)^{\frac{3}{2}}-\frac{2}{3}(1+3 x)^{\frac{1}{2}}+C

Hint: Let 1+3x=t

Given: \int \frac{\left ( 2-3x \right )}{\sqrt{1+3x}}dx

Solution:

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

Let 1+3x=t , 3x=t-1

dx=\frac{dt}{3}
\begin{aligned} &=2 \int(1+3 x)^{\frac{-1}{2}} d x-\frac{1}{3} \int \frac{t-1}{\sqrt{t}} d x \\ &=2\left[\frac{(1+3 x)^{\frac{-1}{2}+1}}{-\frac{1}{2}+1} \times \frac{1}{3}\right]-\frac{1}{3}\left[\int \sqrt{t} d t-\int(t)^{\frac{-1}{2}}\right] \end{aligned}
\begin{aligned} &=\frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{1}{3}\left[2 \times \frac{(t)^{\frac{3}{2}}}{3}-2(t)^{\frac{1}{2}}\right]+C \\ &=\frac{4}{3}(1+3 x)^{\frac{1}{2}}-\frac{2}{9}(1+3 x)^{\frac{3}{2}}-\frac{2}{3}(1+3 x)^{\frac{1}{2}}+C \end{aligned}

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