Step 1: Understanding the Concept:
Integration involving multiple fractional powers of \( x \) (like square roots and cube roots) can be simplified by a rationalizing substitution.
The strategy is to find a variable \( t \) such that every fractional power of \( x \) in the integrand becomes an integer power of \( t \).
This is achieved by setting \( x = t^n \), where \( n \) is the Least Common Multiple (LCM) of all the denominators in the fractional exponents.
In this problem, the powers are \( 1/2 \) and \( 1/3 \), so the denominators are 2 and 3.
Step 2: Key Formula or Approach:
LCM(2, 3) = 6.
Therefore, let \( x = t^6 \).
Differentiating both sides:
\[ dx = 6t^5 \, dt \]
Also, express the terms in the denominator in terms of \( t \):
\[ x^{1/2} = (t^6)^{1/2} = t^3 \]
\[ x^{1/3} = (t^6)^{1/3} = t^2 \]
Step 3: Detailed Explanation:
Substitute these expressions into the original integral:
\[ I = \int \frac{6t^5}{t^3 + t^2} \, dt \]
We can factor out \( t^2 \) from the denominator:
\[ I = 6 \int \frac{t^5}{t^2(t + 1)} \, dt = 6 \int \frac{t^3}{t + 1} \, dt \]
Since the degree of the numerator (3) is greater than the degree of the denominator (1), we perform polynomial division.
A clever shortcut is to add and subtract 1 in the numerator:
\[ \frac{t^3}{t+1} = \frac{t^3 + 1 - 1}{t+1} = \frac{(t+1)(t^2 - t + 1) - 1}{t+1} = (t^2 - t + 1) - \frac{1}{t+1} \]
Now, integrate this simplified expression:
\[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t+1} \right) \, dt \]
\[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \ln |t + 1| \right] + C \]
Multiplying by 6:
\[ I = 2t^3 - 3t^2 + 6t - 6 \ln |t + 1| + C \]
Finally, substitute back \( t = x^{1/6} \):
Since \( t^3 = x^{3/6} = x^{1/2} = \sqrt{x} \) and \( t^2 = x^{2/6} = x^{1/3} = \sqrt[3]{x} \):
\[ I = 2\sqrt{x} - 3\sqrt[3]{x} + 6x^{1/6} - 6 \ln |x^{1/6} + 1| + C \]
Step 4: Final Answer:
The integral evaluates to \( 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6 \ln |x^{1/6} + 1| + C \).
This matches option (B).