Question

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

Hint:

Always use the LCM of the roots for substitution to clear all radical signs at once. Polynomial long division is essential when the numerator's power is equal to or higher than the denominator's power.

Answer 1

Concept:
• When an integrand involves different fractional powers of \(x\), we use a substitution \(x = t^n\) where \(n\) is the least common multiple (LCM) of the denominators of the fractional powers.
• Here, the powers are \(1/2\) and \(1/3\). The LCM of 2 and 3 is 6.
• After substitution, the integral typically converts into a rational function which can be solved by division or partial fractions.

Step 1: Apply substitution to rationalize the denominator
Let \(x = t^6\). Then, differentiate both sides: \[ dx = 6t^5 \, dt \] Substitute \(x\), \(x^{1/2}\), \(x^{1/3}\), and \(dx\) into the integral: \[ x^{1/2} = (t^6)^{1/2} = t^3 \] \[ x^{1/3} = (t^6)^{1/3} = t^2 \] The integral becomes: \[ I = \int \frac{6t^5}{t^3 + t^2} \, dt \]

Step 2: Simplify the rational expression
Factor out \(t^2\) from the denominator: \[ I = \int \frac{6t^5}{t^2(t + 1)} \, dt \] \[ I = 6 \int \frac{t^3}{t + 1} \, dt \]

Step 3: Perform polynomial division
Since the degree of the numerator is greater than the denominator, we divide \(t^3\) by \(t+1\). Recall that \(t^3 + 1 = (t+1)(t^2 - t + 1)\). Thus: \[ t^3 = (t^3 + 1) - 1 = (t+1)(t^2 - t + 1) - 1 \] Dividing by \(t+1\): \[ \frac{t^3}{t+1} = (t^2 - t + 1) - \frac{1}{t+1} \]

Step 4: Integrate the simplified terms
\[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t+1} \right) \, dt \] \[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1| \right] + C \] \[ I = 2t^3 - 3t^2 + 6t - 6\log|t + 1| + C \]

Step 5: Back-substitute for \(x\)
Since \(x = t^6\), then \(t = x^{1/6}\). \[ t^3 = (x^{1/6})^3 = x^{1/2} = \sqrt{x} \] \[ t^2 = (x^{1/6})^2 = x^{1/3} \] The final answer is: \[ I = 2\sqrt{x} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \]