#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.10 Question 3 Maths Textbook Solution.

Answer:  $\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$

Hint: Use substitution method to solve this type of integral

Given:  $\int \frac{x^{2}}{\sqrt{3 x+4}} d x$

Solution:

$I=\int \frac{x^{2}}{\sqrt{3 x+4}} d x$

Put  $3 x+4=t \Rightarrow 3 d x=d t \Rightarrow d x=\frac{d t}{3}$  then

\begin{aligned} &I=\int \frac{\frac{1}{3^{2}}(t-4)^{2}}{\sqrt{t}} \frac{d t}{3} \qquad\left(\because x=\frac{t-4}{3}\right) \\ & \end{aligned}

$\Rightarrow I=\frac{1}{3 \times 3^{2}} \int \frac{t^{2}+16-8 t}{\sqrt{t}} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left\{\frac{t^{2}}{\sqrt{t}}+16 \frac{1}{\sqrt{t}}-8 \cdot \frac{t}{\sqrt{t}}\right\} d t \\ & \end{aligned}

$\Rightarrow I=\frac{1}{27} \int\left\{t^{2-\frac{1}{2}}+16 t^{-\frac{1}{2}}-8 t^{1-\frac{1}{2}}\right\} d t$

\begin{aligned} &\Rightarrow I=\frac{1}{27} \int\left(t^{\frac{3}{2}}-8 t^{\frac{1}{2}}+16 t^{-\frac{1}{2}}\right) d t \\ & \end{aligned}

$\Rightarrow I=\frac{1}{27}\left[\int t^{\frac{3}{2}} d t-8 \int t^{\frac{1}{2}} d t+16 \int t^{-\frac{1}{2}} d t\right]$

$\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-8 \cdot \frac{t^{\frac{1}{2}}{2}+1} + 16 \cdot \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\right] \qquad\left[\because f t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$

\begin{aligned} &\Rightarrow I=\frac{1}{27}\left[\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-8 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+16 \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\right] \\ & \end{aligned}

$\Rightarrow I=\frac{1}{27}\left[\frac{2}{5}(t)^{\frac{5}{2}}-8 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+16.2(t)^{\frac{1}{2}}+c\right] \\$

$\Rightarrow I=\frac{2}{135}(3 x+4)^{\frac{5}{2}}-\frac{16}{81}(3 x+4)^{\frac{3}{2}}+\frac{32}{27}(3 x+4)^{\frac{1}{2}}+c$