Question

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

• 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

Let \( x = t^4 \), so \( dx = 4t^3 dt \). Substituting into the integral gives \[f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx = \int \frac{4t^3}{t(1 + t)} \, dt\]. This simplifies to \[f(x) = 4 \int \frac{t^2 - 1 + 1}{1 + t} \, dt = 4 \int \frac{t^2 - 1}{1 + t} \, dt + 4 \int \frac{1}{1 + t} \, dt\]. Further breaking it down yields \[f(x) = 4 \left( \int (t-1) \, dt + \int \frac{1}{1 + t} \, dt \right)\]. This results in \[f(x) = 4 \left( \frac{(t-1)^2}{2} + \ln(1 + t) + C \right)\]. Since \( t = x^{1/4} \), the final expression for \( f(x) \) is \[f(x) = 2 \left( x^{1/4} - 1 \right)^2 + 4 \ln(1 + x^{1/4}) + C\]. Given \( f(0) = -6 \), we solve for \( C \): \[f(0) = 2 \times (0^{1/4} - 1)^2 + 4 \ln(1 + 0^{1/4}) + C = -6 \quad \Rightarrow \quad C = -8\]. Now, for \( f(1) \): \[f(1) = 4 \ln 2 - 8 = 4(\ln 2 - 2)\].