∫ fracdx(x^2+4)^3/2 is equal to ( where C is an arbitrary constant)

The correct answer is Option (1) → \( \displaystyle\frac{x}{4\sqrt{x^2+4}}+C \)

The form \( x^2+a^2 \) suggests the substitution \( x=2\tan\theta \). Then

    \( dx=2\sec^2\theta\,d\theta \)

    \( (x^2+4)^{3/2}=(4\sec^2\theta)^{3/2}=8\sec^3\theta \)

So

    \( \displaystyle \int\frac{dx}{(x^2+4)^{3/2}}=\int\frac{2\sec^2\theta}{8\sec^3\theta}\,d\theta \)

    \( \displaystyle =\frac{1}{4}\int\cos\theta\,d\theta \)

    \( \displaystyle =\frac{1}{4}\sin\theta+C \)

Since \( \tan\theta=\frac{x}{2} \), \( \sin\theta=\frac{x}{\sqrt{x^2+4}} \). So

    \( \displaystyle \int\frac{dx}{(x^2+4)^{3/2}}=\frac{x}{4\sqrt{x^2+4}}+C \)

Shortcut: Differentiate the expression in Option (1):

    \( \displaystyle \frac{d}{dx}\left[\frac{x}{4\sqrt{x^2+4}}\right]=\frac{1}{4}\cdot\frac{\sqrt{x^2+4}-\frac{x^2}{\sqrt{x^2+4}}}{x^2+4} \)

    \( \displaystyle =\frac{1}{4}\cdot\frac{(x^2+4)-x^2}{(x^2+4)^{3/2}} \)

    \( \displaystyle =\frac{1}{(x^2+4)^{3/2}} \)

This is the integrand, so Option (1) is correct.