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: Understanding the Question:
The problem requires finding the antiderivative of a rational function involving fractional exponents ($\frac{1}{2}$ and $\frac{1}{3}$) of $x$.

Step 2: Key Formula or Approach:
To eliminate fractional exponents, we use a substitution of the form $x = t^n$, where $n$ is the least common multiple (LCM) of the denominators of the fractional exponents.
The LCM of $2$ and $3$ is $6$. Therefore, we substitute:
$$x = t^6 \implies dx = 6t^5\ dt$$ This substitution turns the fractional exponents into standard integer powers.

Step 3: Detailed Explanation:
Let the given integral be $I$. Applying the substitution components:
$$x^{\frac{1}{2}} = (t^6)^{\frac{1}{2}} = t^3$$ $$x^{\frac{1}{3}} = (t^6)^{\frac{1}{3}} = t^2$$ Substitute these variables into the integral expression:
$$I = \int \frac{6t^5}{t^3 + t^2}\ dt$$ Factor out $t^2$ from the denominator:
$$I = \int \frac{6t^5}{t^2(t + 1)}\ dt = 6 \int \frac{t^3}{t + 1}\ dt$$ To solve this algebraic fraction, add and subtract $1$ in the numerator to execute a polynomial long division shortcut:
$$I = 6 \int \frac{(t^3 + 1) - 1}{t + 1}\ dt$$ Using the algebraic identity $t^3 + 1 = (t + 1)(t^2 - t + 1)$, we split the fraction:
$$I = 6 \int \frac{(t + 1)(t^2 - t + 1)}{t + 1}\ dt - 6 \int \frac{1}{t + 1}\ dt$$ $$I = 6 \int (t^2 - t + 1)\ dt - 6 \int \frac{1}{t + 1}\ dt$$ Integrate each term individually with respect to $t$:
$$I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t \right] - 6\log|t + 1| + c$$ $$I = 2t^3 - 3t^2 + 6t - 6\log|t + 1| + c$$ Now, substitute back $t = x^{\frac{1}{6}}$, which means $t^2 = x^{\frac{1}{3}}$ and $t^3 = x^{\frac{1}{2}}$:
$$I = 2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\log|x^{\frac{1}{6}} + 1| + c$$

Step 4: Final Answer:
The evaluated integral matches option (B).