Question

If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6, { then } f(1) { is equal to:} \]

• A. \( \log 2 + 2 \)
• B. \( 4 (\log 2 - 2) \)
• C. \( 2 - \log 2 \)
• D. \( 4 (\log 2 + 2) \)

Hint:

For integrals involving powers of \( x \), substitution can simplify the expression and make the integrals easier to evaluate.

Answer 1

The Correct Option is A

Step 1: Substitution in the Integral

Let \( x = t^4 \), so that \( dx = 4t^3 dt \).
Substituting this into the given integral: \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx = \int \frac{4t^3}{t(1 + t)} \, dt \] Simplifying, \[ f(x) = 4 \int \frac{t^2 - 1 + 1}{1 + t} \, dt \]

Step 2: Break the Integral into Parts

Expanding further: \[ f(x) = 4 \int (t-1) \, dt + 4 \int \frac{1}{1 + t} \, dt \] Evaluating each part: \[ f(x) = 4 \left( \frac{(t-1)^2}{2} + \ln(1 + t) + C \right) \]

Step 3: Back Substitution

Since \( t = x^{1/4} \), the final expression for \( f(x) \) is: \[ f(x) = 2 \left( x^{1/4} - 1 \right)^2 + 4 \ln(1 + x^{1/4}) + C \]

Step 4: Solving for \( C \)

Given \( f(0) = -6 \), \[ f(0) = 2 \times (0^{1/4} - 1)^2 + 4 \ln(1 + 0^{1/4}) + C \] Evaluating, \[ f(0) = 2 \times 1 + 4 \times 0 + C = -6 \] Solving for \( C \): \[ C = -8 \]

Step 5: Compute \( f(1) \)

Now, \[ f(1) = 2 \times (1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8 \] Simplifying, \[ f(1) = 4 \ln 2 - 8 = 4 (\ln 2 - 2) \]

Final Answer: \( 4(\ln 2 - 2) \)