Question

$\int \left[\frac{f\text{'}\left(x\right)}{\sqrt{f\left(x\right)}}\right]dx=\_\_\_\_\_\_\_+c$; $f\left(x\right)\ne 0$

• A. $\left(\frac{1}{2}\right)\sqrt{f\left(x\right)}$
• B. $2\sqrt{f\left(x\right)}$
• C. $\left(\frac{1}{2}\right)f\left(x\right)$
• D. $2f\left(x\right)$

Answer 1

The correct option is B

$2\sqrt{f\left(x\right)}$

Explanation for the correct option:

Finding the value of $\int \left[\frac{f\text{'}\left(x\right)}{\sqrt{f\left(x\right)}}\right]dx$

Consider the given Equation as,

$I=\int \left[\frac{f\text{'}\left(x\right)}{\sqrt{f\left(x\right)}}\right]dx$

Let,

$$\begin{gathered} f\left(x\right)=t \\ f\text{'}\left(x\right)dx=dt \end{gathered}$$ [differentiating]

Then $I$ becomes,

$$\begin{gathered} I=\int \left[\frac{dt}{\sqrt{t}}\right] \\ I=\int t^{-\frac{1}{2}}dt \end{gathered}$$

We know that

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$

Then $I$ becomes,

$$\begin{gathered} I=\int t^{-\frac{1}{2}}dt \\ =\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}+c \\ =\frac{t^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}+c \\ I=2t^{\frac{1}{2}}+c \end{gathered}$$

Substitute $t=f\left(x\right)$ then the above equation becomes,

$$\begin{gathered} I=2{\left(f\left(x\right)\right)}^{\frac{1}{2}}+c \\ \Rightarrow I=2\sqrt{f\left(x\right)}+c \end{gathered}$$

Hence , the correct answer is Option (B).