Question

Find : \( \int \frac{dx}{x^{1/2} + x^{1/3}} \)

Hint:

Adding and subtracting 1 inside the numerator polynomial (\( t^3 \to t^3+1-1 \)) allows you to use the sum of cubes factorization formula \( a^3+b^3 = (a+b)(a^2-ab+b^2) \), avoiding long polynomial division entirely.

Answer 1

Concept: When an integrand involves fractional powers of \( x \), we can eliminate the radical symbols completely by selecting a substitution variable power equal to the least common multiple (LCM) of the denominators of the fractional exponents. Here, the exponents are \( \frac{1}{2} \) and \( \frac{1}{3} \). The LCM of 2 and 3 is 6.

Step 1: Set up the algebraic substitution.
Let: \[ x = t^6 \quad \Rightarrow \quad dx = 6t^5 \, dt \] Now find expressions for the terms inside the denominator: \[ x^{1/2} = (t^6)^{1/2} = t^3 \] \[ x^{1/3} = (t^6)^{1/3} = t^2 \]

Step 2: Substitute components into the integral and simplify.
Substituting these values back into the expression: \[ I = \int \frac{6t^5 \, dt}{t^3 + t^2} \] Factor out \( t^2 \) from the denominator to cancel terms with the numerator: \[ I = \int \frac{6t^5}{t^2(t + 1)} \, dt = 6 \int \frac{t^3}{t + 1} \, dt \]

Step 3: Perform polynomial division on the integrand.
To integrate a rational expression where the numerator's degree is higher than the denominator, apply algebraic division or use a clever manipulation: \[ t^3 = (t^3 + 1) - 1 = (t + 1)(t^2 - t + 1) - 1 \] Dividing each part by \( (t + 1) \): \[ \frac{t^3}{t + 1} = \frac{(t + 1)(t^2 - t + 1) - 1}{t + 1} = t^2 - t + 1 - \frac{1}{t + 1} \]

Step 4: Integrate the terms step-by-step.
Substitute the broken down terms back into the integral: \[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t + 1} \right) dt \] \[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1| \right] + C \] Distributing the constant factor 6: \[ I = 2t^3 - 3t^2 + 6t - 6\log|t + 1| + C \]

Step 5: Back-substitute original variable parameters.
Since \( x = t^6 \), we have \( t = x^{1/6} \). Let's convert high powers back:
• \( t^3 = (x^{1/6})^3 = x^{1/2} \)
• \( t^2 = (x^{1/6})^2 = x^{1/3} \)
• \( t = x^{1/6} \) Substituting these back yields the final solution: \[ I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \]