Question

$\int \frac{dx}{x^2\left(\sqrt{4-x^2}\right)}=$

• A. $\left(\frac{1}{4}\right)\left(\frac{\sqrt{4–x^2}}{x}\right)+c$
• B. $\left(\frac{1}{2}\right)\left(\frac{\sqrt{4–x^2}}{x}\right)+c$
• C. $\left(\frac{-1}{4}\right)\left(\frac{\sqrt{4–x^2}}{x}\right)+c$
• D. $\left(\frac{-1}{2}\right)\left(\frac{\sqrt{4–x^2}}{x}\right)+c$

Answer 1

The correct option is C

$\left(\frac{-1}{4}\right)\left(\frac{\sqrt{4–x^2}}{x}\right)+c$

Explanation for the correct option:

Finding the integral:

$\int \frac{dx}{x^2\left(\sqrt{4-x^2}\right)}$

$$\begin{gathered} \begin{array}{l}=\int \frac{dx}{x^2\sqrt{x^2\left(\frac{4}{x^2}-1\right)}}\end{array} \\ =\int \frac{dx}{x^3\sqrt{\left(\frac{4}{x^2}-1\right)}}\left[\because \sqrt{x^2}=x\right] \end{gathered}$$

Using the substitution method,

$$\begin{gathered} t=\sqrt{\frac{4}{x^2}-1} \\ {t}^2=\frac{4}{x^2}-1 \\ {t}^2=4x^{-2}-1 \end{gathered}$$

Differentiating with respect to $t$

$$\begin{gathered} 2tdt=\frac{(-2)4}{x^3}dx \\ tdt=\frac{-4}{x^3}dx \\ \frac{dx}{x^3}=\frac{-1}{4}tdt \end{gathered}$$

Substituting the values, we get

$$\begin{gathered} =\frac{-1}{4}\int \frac{tdt}{t} \\ =\frac{-1}{4}t+c \\ =\frac{-1}{4}\sqrt{\frac{4-x^2}{x}}+c \end{gathered}$$

Hence the correct answer is option (C).