Question

Evaluate the indefinite integral:
$$\int \frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}\ dx$$

• A. $x - x^{\frac{1}{3}} + x^{\frac{1}{6}} - \log|x^{\frac{1}{6}} + 1| + c$
• B. $2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\log|x^{\frac{1}{6}} + 1| + c$
• C. $2x^{\frac{1}{2}} + 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} + 6\log|x^{\frac{1}{6}} + 1| + c$
• D. $x + x^{\frac{1}{3}} + x^{\frac{1}{6}} + \log|x^{\frac{1}{6}} + 1| + c$

Hint:

When an integrand contains multiple radical expressions like $\sqrt[p]{x}$ and $\sqrt[q]{x}$, your first step should always be choosing $x = t^{\text{LCM}(p,q)}$. This reliably transforms any radical algebraic fraction into a standard polynomial expression.

Answer 1

The Correct Option is B

Step 1: See the problem.
We integrate $\displaystyle \int \frac{1}{x^{1/2} + x^{1/3}}\,dx$. The fractional powers $\tfrac12$ and $\tfrac13$ make it messy, so we remove them with a substitution.
Step 2: Choose the substitution.
Take the LCM of the denominators 2 and 3, which is 6. Let $x = t^6$, so $dx = 6t^5\,dt$. This makes the powers whole numbers.
Step 3: Rewrite each piece.
$x^{1/2} = t^3$ and $x^{1/3} = t^2$. The integral becomes \[ I = \int \frac{6t^5}{t^3 + t^2}\,dt \]
Step 4: Simplify the fraction.
Take $t^2$ out of the bottom: $t^3 + t^2 = t^2(t+1)$, so \[ I = 6\int \frac{t^3}{t+1}\,dt \]
Step 5: Do the division.
Write $t^3 = (t^3+1) - 1$ and use $t^3+1 = (t+1)(t^2-t+1)$: \[ I = 6\int (t^2 - t + 1)\,dt - 6\int \frac{1}{t+1}\,dt \] \[ = 6\left(\frac{t^3}{3} - \frac{t^2}{2} + t\right) - 6\ln|t+1| + c = 2t^3 - 3t^2 + 6t - 6\ln|t+1| + c \]
Step 6: Put $x$ back.
Since $t = x^{1/6}$, we have $t^3 = x^{1/2}$, $t^2 = x^{1/3}$, $t = x^{1/6}$: \[ I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\ln|x^{1/6} + 1| + c \] \[ \boxed{2x^{\frac12} - 3x^{\frac13} + 6x^{\frac16} - 6\log|x^{\frac16} + 1| + c} \]