Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.10 Question 2 Maths Textbook Solution.

Answer:  $\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c$

Hint: Use substitution method to solve this type of integral

Given:  $\int \frac{x^{2}}{\sqrt{x-1}} d x$

Solution:

Let  $I=\int \frac{x^{2}}{\sqrt{x-1}} d x$

Substitute $x-1=t \Rightarrow d x=d t$  then

\begin{aligned} I &=\int \frac{(t+1)^{2}}{\sqrt{t}} d t \qquad(\because x=t+1) \\ & \end{aligned}

$=\int \frac{t^{2}+1+2 t}{\sqrt{t}} d t \qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\$

$=\int\left(\frac{t^{2}}{\sqrt{t}}+\frac{1}{\sqrt{t}}+2 \frac{t}{\sqrt{t}}\right) d t \\$

$=\int\left(t^{2-\frac{1}{2}}+t^{-\frac{1}{2}}+2 t^{1-\frac{1}{2}}\right) d t$

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

$=\int t^{\frac{3}{2}} d t+\int t^{-\frac{1}{2}} d t+2 \int t^{\frac{1}{2}} d t$

\begin{aligned} &=\frac{t^{\frac{3}{2}}{2}}{\frac{3}{2}+1}+\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c \qquad\left[\because f^{n} d t=\frac{t^{n+1}}{n+1}+c\right] \\ & \end{aligned}

$=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c$

\begin{aligned} &=\frac{2}{5}(t)^{\frac{5}{2}}+2(t)^{\frac{1}{2}}+2 \cdot \frac{2}{3}(t)^{\frac{3}{2}}+c \\ & \end{aligned}

$=\frac{2}{5}(x-1)^{\frac{5}{2}}+2(x-1)^{\frac{1}{2}}+\frac{4}{3}(x-1)^{\frac{3}{2}}+c \qquad[\because x-1=t]$