Substitution: Start by substituting $ x = t^4 $ which implies $ dx = 4t^3 \, dt $.

Integral Transformation:

$ \int \frac{1}{x^{1/4}(1 + x^{1/4})} \, dx \Rightarrow \int \frac{4t^3 \, dt}{t(1 + t)} = \int \frac{4t}{1 + t} \, dt $

Rewriting the Integral: The expression $ \frac{4t}{1 + t} $ can be decomposed as:

$ \frac{4t}{1 + t} = 4\left(\frac{t^2 - 1 + 1}{1 + t}\right) = 4\left((t - 1) + \frac{1}{t + 1}\right) $

Separate and Integrate:

$ 4 \int (t - 1) \, dt + 4 \int \frac{1}{t + 1} \, dt $

Solving the Integrals:

The integral of $ t - 1 $ is $ \frac{(t - 1)^2}{2} $.

The integral of $ \frac{1}{t + 1} $ is $ \ln(t + 1) $.

Therefore:

$ f(x) = 4 \left\{ \frac{(t - 1)^2}{2} + \ln(t + 1) \right\} + C $

Substitute $ t = x^{1/4} $: Replace $ t $ back with $ x^{1/4} $:

$ f(x) = 2(x^{1/4} - 1)^2 + 4 \ln(1 + x^{1/4}) + C $

Using the Condition $ f(0) = -6 $:

$ f(0) = 2(0 - 1)^2 + 4 \ln(1 + 0) + C = 2 \times 1 - 0 + C = -6 $

Solving gives $ C = -8 $.

Find $ f(1) $:

$ f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8 $

Simplifies to:

$ f(1) = 0 + 4 \ln 2 - 8 = 4(\ln 2 - 2) $