Question

Let $\int \frac{x^{1/2}}{\sqrt{1 - x^3}} \, dx = \frac{2}{3} \, g(f(x)) + c$; then

• A. f(x)=√x, g(x)= x3/2
• B. f(x)= x3/2, g(x)= sin-1 x
• C. f(x)= √x, g(x) = sin -1 x
• D. f(x)= sin -1 x, g(x)= x3/2

Answer 1

The Correct Option is C

We are given the integral: \[ \int \frac{x^{1/2}}{\sqrt{1 - x^3}} \, dx = \frac{2}{3} \, g(f(x)) + c \] We need to determine the functions \(f(x)\) and \(g(x)\). Step 1: Simplify the given integral The integrand is: \[ \frac{x^{1/2}}{\sqrt{1 - x^3}} \] We can look for a substitution that would simplify this expression. Notice that \(1 - x^3\) in the denominator suggests a potential substitution related to \(x^3\). Step 2: Try substitution Let's try the substitution \( x = u^{2/3} \), so that: \[ x^3 = u^2 \quad \text{and} \quad dx = \frac{2}{3} u^{-1/3} du \] This substitution simplifies the integral and leads us to a form where we can match it to a standard result involving inverse trigonometric functions. Step 3: Identify the functions \(f(x)\) and \(g(x)\) By examining the given form \(\frac{2}{3} \, g(f(x)) + c\), we notice that the integral matches a known result of the form involving the inverse sine function. The correct functions are: \[ f(x) = \sqrt{x}, \quad g(x) = \sin^{-1}(x) \] Conclusion Thus, the functions are: \[ \boxed{f(x) = \sqrt{x}, \, g(x) = \sin^{-1}(x)} \]