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

Answer:  $\frac{2}{27}(6 x+1) \sqrt{3 x+2}+C$

Hint: Use Integration by partial function.

Given: $\int \frac{2x+1}{\sqrt{3x+2}}dx$

Solution: On multiplying and dividing by 3 in the equation, we get

$=\frac{1}{3} \int\left(\frac{6 x+3}{\sqrt{3 x+2}}\right) d x$

The equation can be written as

$=\frac{1}{3} \int\left(\frac{6 x+4-1}{\sqrt{3 x+2}}\right) d x$

So taking 2 common from and subtracting

$=\frac{1}{3} \int\left(\frac{2(3 x+2)}{\sqrt{3 x+2}}-\frac{1}{\sqrt{3 x+2}}\right) d x$

On simplifying

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

By splitting the integral

$=\frac{1}{3} \int 2(3 x+2)^{\frac{1}{2}} d x-\frac{1}{3}\int(3 x+2)^{-\frac{1}{2}} d x$

So integrating we get

\begin{aligned} &=\frac{1}{3}\left[2\left\{\frac{(3 x+2)^{\frac{1}{2}+1}}{3\left(\frac{1}{2}+1\right)}\right\}-\frac{(3 x+2)^{-\frac{1}{2}+1}}{3\left(-\frac{1}{2}+1\right)}\right]+C \\ &=\frac{1}{3}\left[\frac{4}{9}(3 x+2)^{\frac{3}{2}}-\frac{2}{3}(3 x+2)^{\frac{1}{2}}\right]+C \end{aligned}

By Simplifying, we get

\begin{aligned} &=\frac{4}{27}(3 x+2)^{\frac{3}{2}}-\frac{2}{9}(3 x+2)^{\frac{1}{2}}+C \\ &=\sqrt{3 x+2}\left[\frac{4}{27}(3 x+2)-\frac{2}{9}\right]+C \\ &=\sqrt{3 x+2}\left[\frac{4(3 x+2)-6}{27}\right]+C \\ &=\sqrt{3 x+2}\left[\frac{12 x+8-6}{27}\right]+C \\ &=\frac{2}{27}(6 x+1) \sqrt{3 x+2}+C \end{aligned}