#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.10 question 5

Answer:  $\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$

Hint: Use substitution method to solve this type of integral

Given:  $\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$

Solution: Let  $I=\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$

Put  $x+2=t \Rightarrow d x=d t$

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

$\Rightarrow I=\int\left\{2\left(t^{2}-4 t+4\right)+3\right\} \sqrt{t} d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

\begin{aligned} &\Rightarrow I=\int\left\{2 t^{2}-8 t+8+3\right\} \sqrt{t} d t \\ & \end{aligned}

$\Rightarrow I=\int\left(2 t^{2}-8 t+11\right) \sqrt{t} d t \\$

$\Rightarrow I=\int\left(2 t^{2} \sqrt{t}-8 t \cdot \sqrt{t}+11 \sqrt{t}\right) d t$

\begin{aligned} &\Rightarrow I=\int\left(2 t^{2+\frac{1}{2}}-8 t^{1+\frac{1}{2}}+11 t^{\frac{1}{2}}\right) d t \\ & \end{aligned}

$\Rightarrow I=\left[2 \int t^{\frac{5}{2}} d t-8 \int t^{\frac{3}{2}} d t+11 \int t^{\frac{1}{2}} d t\right]$

$\Rightarrow I=\left[2 \cdot \frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}-8 \cdot \frac{t^{\frac{3}{2}}{ } \frac{1}{2}+1} + 11 \cdot \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]$

\begin{aligned} &\Rightarrow I=\left[2 \cdot \frac{t^{\frac{7}{2}}}{\frac{7}{2}}-8 \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}}+11 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}

$\Rightarrow I=2 \cdot \frac{2}{7} t^{\frac{7}{2}}-8 \cdot \frac{2}{5} t^{\frac{5}{2}}+11 \cdot \frac{2}{3} t^{\frac{3}{2}}+c \\$

$\Rightarrow I=\frac{4}{7}(x+2)^{\frac{7}{2}}-\frac{16}{5}(x+2)^{\frac{5}{2}}+\frac{22}{3}(x+2)^{\frac{3}{2}}+c$