Question

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

• A. \(4(\ln2-2)\)
• B. \(4(\ln2-1)\)
• C. \(2(\ln2-1)\)
• D. \(6(\ln2-2)\)

Hint:

Whenever fractional exponents appear, try substitution \(x=t^n\) to convert everything into polynomials.

Answer 1

The Correct Option is A

Concept: Fractional powers in integrals are best handled using substitution \(x=t^4\), which removes roots and converts the expression into a rational function.

Step 1: Substitute \(x=t^4\). \[ dx=4t^3dt \] \[ x^{1/4}=t \] So integral becomes: \[ f(x)=\int \frac{4t^3}{t(1+t)}dt \] \[ =4\int \frac{t^2}{1+t}dt \]

Step 2: Perform algebraic division. \[ \frac{t^2}{1+t}=t-1+\frac{1}{1+t} \] So: \[ f(x)=4\int\left(t-1+\frac{1}{1+t}\right)dt \]

Step 3: Integrate term by term. \[ f(x)=4\left(\frac{t^2}{2}-t+\ln(1+t)\right)+C \]

Step 4: Use condition \(f(0)=-6\). At \(x=0\Rightarrow t=0\): \[ C=-6 \]

Step 5: Evaluate at \(x=1\). At \(x=1\Rightarrow t=1\): \[ f(1)=4\left(\frac12-1+\ln2\right)-6 \] \[ =4(\ln2-2) \] \[ \boxed{4(\ln2-2)} \]