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. \( 4 (\log 2 - 2) \)
• B. \( \log 2 + 2 \)
• 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

We are given the integral and asked to solve for \( f(x) \) in a detailed step-by-step manner.

Step 1: Substitution 

The given integral is: \[ \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx \] Let \( x = t^4 \). Then, the differential \( dx \) becomes: \[ dx = 4t^3 \, dt \] Substituting into the integral: \[ \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx = \int \frac{4t^3}{t(1 + t)} \, dt \]

Step 2: Simplifying the Expression

The integral simplifies to: \[ \int \frac{4t^3}{t(1 + t)} \, dt = \int 4 \frac{t^2}{1+t} \, dt \] We now split the integral: \[ 4 \int \frac{t^2}{1 + t} \, dt \] Perform polynomial division on \( \frac{t^2}{1+t} \) to get: \[ \frac{t^2}{1+t} = t - 1 + \frac{1}{1+t} \] So, the integral becomes: \[ 4 \int \left( t - 1 + \frac{1}{1+t} \right) \, dt \]

Step 3: Integrating Each Term

We now integrate each term: \[ \int t \, dt = \frac{t^2}{2}, \quad \int 1 \, dt = t, \quad \int \frac{1}{1+t} \, dt = \ln(1+t) \] Therefore, the integral becomes: \[ f(x) = 4 \left( \frac{t^2}{2} - t + \ln(1+t) \right) + C \] Substituting back \( t = x^{1/4} \) into the expression, we get: \[ f(x) = 4 \left( \frac{x^{1/2}}{2} - x^{1/4} + \ln(1 + x^{1/4}) \right) + C \]

Step 4: Using the Initial Condition

We are given that \( f(0) = -6 \). Substituting \( x = 0 \) into the equation: \[ f(0) = 4 \left( \frac{0^{1/2}}{2} - 0^{1/4} + \ln(1 + 0^{1/4}) \right) + C = C \] Therefore, \( C = -6 \). The function becomes: \[ f(x) = 4 \left( \frac{x^{1/2}}{2} - x^{1/4} + \ln(1 + x^{1/4}) \right) - 6 \]

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

Now, substitute \( x = 1 \) into the expression for \( f(x) \): \[ f(1) = 4 \left( \frac{1^{1/2}}{2} - 1^{1/4} + \ln(1 + 1^{1/4}) \right) - 6 \] Simplifying: \[ f(1) = 4 \left( \frac{1}{2} - 1 + \ln(2) \right) - 6 \] \[ f(1) = 4 \left( -\frac{1}{2} + \ln(2) \right) - 6 \] \[ f(1) = -2 + 4 \ln(2) - 6 = -8 + 4 \ln(2) \] Thus, the final expression for \( f(1) \) is: \[ f(1) = 4 \ln(2) - 8 = 4 (\ln 2 - 2) \]

Final Answer:

\[ f(1) = 4 (\ln 2 - 2) \] Therefore, the correct option is \( 4 (\log_e 2 - 2) \).

Answer 2

Step 1: Write the given function.
We are given: \[ f(x) = \int \frac{1}{x^{1/4}(1 + x^{1/4})} \, dx, \quad f(0) = -6 \]

Step 2: Substitution to simplify the integral.
Let: \[ t = x^{1/4} \Rightarrow x = t^4 \Rightarrow dx = 4t^3 \, dt \] Substitute in the integral: \[ f(x) = \int \frac{1}{t(1 + t)} \cdot 4t^3 \, dt = 4 \int \frac{t^2}{1 + t} \, dt \]

Step 3: Simplify the integrand.
Perform division: \[ \frac{t^2}{1 + t} = t - 1 + \frac{1}{1 + t} \] Thus: \[ f(x) = 4 \int \left( t - 1 + \frac{1}{1 + t} \right) dt \]

Step 4: Integrate term by term.
\[ f(x) = 4 \left( \frac{t^2}{2} - t + \log|1 + t| \right) + C \] Simplify: \[ f(x) = 2t^2 - 4t + 4\log(1 + t) + C \] Now, substitute back \( t = x^{1/4} \): \[ f(x) = 2x^{1/2} - 4x^{1/4} + 4\log(1 + x^{1/4}) + C \]

Step 5: Apply initial condition \( f(0) = -6 \).
\[ f(0) = 2(0) - 4(0) + 4\log(1 + 0) + C = C = -6 \] So: \[ f(x) = 2x^{1/2} - 4x^{1/4} + 4\log(1 + x^{1/4}) - 6 \]

Step 6: Find \( f(1) \).
Substitute \( x = 1 \): \[ f(1) = 2(1) - 4(1) + 4\log(1 + 1) - 6 \] \[ f(1) = 2 - 4 + 4\log(2) - 6 = 4(\log 2 - 2) \]

Final Answer:
\[ \boxed{f(1) = 4(\log 2 - 2)} \]