Question

Let \( f(x) = \int \frac{(2 - x^2) \cdot e^x{(\sqrt{1 + x})(1 - x)^{3/2}} dx \). If \( f(0) = 0 \), then \( f\left(\frac{1}{2}\right) \) is equal to :}

• A. \( \sqrt{2e} - 1 \)
• B. \( \sqrt{3e} - 1 \)
• C. \( \sqrt{3e} + 1 \)
• D. \( \sqrt{2e} + 1 \)

Hint:

Whenever you see \( e^x \) in an algebraic integral, immediately check if the remaining part can be split into a function and its derivative.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves indefinite integration of a function containing \( e^x \). Usually, such integrals follow the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \). We need to manipulate the integrand to fit this structure.
Step 2: Key Formula or Approach:
1. Recognize the denominator: \( (\sqrt{1+x})(1-x)\sqrt{1-x} = (1-x^2)^{1/2}(1-x) \).
2. Use the property \( \int e^x [g(x) + g'(x)] dx = e^x g(x) \).
Step 3: Detailed Explanation:
The integrand can be written as: \[ I = \int e^x \left[ \frac{2 - x^2}{(1 - x^2)^{1/2}(1 - x)} \right] dx \] Let's test \( g(x) = \sqrt{\frac{1+x}{1-x}} \). Then \( g'(x) = \frac{1}{2\sqrt{\frac{1+x}{1-x}}} \cdot \frac{(1-x) - (1+x)(-1)}{(1-x)^2} = \frac{1}{(1-x)\sqrt{1-x^2}} \). By splitting the numerator \( 2-x^2 = (1-x^2) + 1 \), we can show the integrand is exactly \( e^x (g(x) + g'(x)) \). Thus, \( f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C \). Given \( f(0) = 0 \): \[ e^0 \sqrt{\frac{1+0}{1-0}} + C = 0 \implies 1 + C = 0 \implies C = -1 \] Now find \( f(1/2) \): \[ f(1/2) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} - 1 = \sqrt{e} \sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{3e} - 1 \] Step 4: Final Answer:
The value is \( \sqrt{3e} - 1 \).