#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.10 Question 9

Answer:  $2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$

Hint: Use substitution method to solve this type of integral

Given:  $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$

Solution:

Let  $I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$

In the given integral, the exponent of  $x \text { are } \frac{1}{2} \& \frac{1}{4}$  and the LCM of denominators is 4

Substitute  $x=t^{4} \Rightarrow d x=4 t^{3} d t$   then

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

$\Rightarrow I=4 \int \frac{t^{3}}{t^{2}+t} d t \\$

$\Rightarrow I=\int \frac{t^{3}+t^{2}-t^{2}}{t^{2}+t} d t\qquad\left[\text { we can write } t^{3}=t^{3}+t^{2}-t^{2}\right]$

\begin{aligned} &\Rightarrow I=4 \int \frac{t\left(t^{2}+t\right)-t^{2}}{\left(t^{2}+t\right)} d t \\ & \end{aligned}

$\Rightarrow I=4 \int\left(t-\frac{t^{2}}{t^{2}+t}\right) d t \\$

$\Rightarrow I=4 \int\left\{t-\frac{t^{2}+t}{t^{2}+t}-\frac{t}{t^{2}+t}\right\} d t$

\begin{aligned} &\Rightarrow I=4 \int\left\{t-1-\frac{t}{t(t+1)}\right\} d t=4 \int\left\{t-1-\frac{1}{(t+1)}\right\} d t \\ & \end{aligned}

$\Rightarrow I=4 \int t d t-4 \int t^{0} d t-4 \int \frac{1}{t+1} d t$

\begin{aligned} &\Rightarrow I=4 \frac{t^{1+1}}{1+1}-4 \log |t+1|-4 t+c \\ & \end{aligned}

$\Rightarrow I=4 \frac{t^{2}}{2}-4 t-4 \log |t+1|+c \\$

$\therefore I=2 \sqrt{x}-4 x^{\frac{1}{4}}+4 \log \left|1+x^{\frac{1}{4}}\right|+c$