#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.10 Question 1 Maths Textbook Solution.

Answer:  $\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c$

Hint: To solve this type of problem, we use substitution method. Let  $x+2=t$  then solve the integral by general  method.

Given:  $\int x^{2} \sqrt{x+2} d x$

Solution:

Let  $I=\int x^{2} \sqrt{x+2} d x$

Substitute  $x+2=t \Rightarrow d x=d t$

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

$=\int\left(t^{2}+4-4 t\right) t^{\frac{1}{2}} d t \qquad\qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \\$

$=\int\left(t^{2} \cdot t^{\frac{1}{2}}+4 t^{\frac{1}{2}}-4 t . t^{\frac{1}{2}}\right) d t \\$

$=\int\left(t^{\frac{5}{2}}+4 t^{\frac{1}{2}}-4 t^{\frac{3}{2}}\right) d t \qquad\qquad\left(\because a^{m} \cdot a^{n}=a^{m+n}\right)$

$\begin{array}{r} =\int t^{\frac{5}{2}} d t+4 \int t^{\frac{1}{2}} d t-4 \int t^{\frac{3}{2}} d t \\ \end{array}$

$\quad\left[\because \int\{f(x) \pm a g(x)\} d x=\int f(x) d x \pm a \int g(x) d x\right]$

$=\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+4 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-4 \frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c \qquad\qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$

\begin{aligned} &=\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+4 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4 \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+c \\ & \end{aligned}

$=\frac{2}{7}(t)^{\frac{7}{2}}-4 \cdot \frac{2}{5}(t)^{\frac{5}{2}}+4 \frac{2}{3}(t)^{\frac{3}{2}}+c \\$

$=\frac{2}{7}(x+2)^{\frac{7}{2}}-\frac{8}{5}(x+2)^{\frac{5}{2}}+\frac{8}{3}(x+2)^{\frac{3}{2}}+c \qquad \qquad[\because x+2=t]$