Evaluate the integral: I = integral (sqrt(x) - 2 cube
Evaluate the integral: $$ I = \int \frac{\sqrt{x} - 2 \sqrt[3]{x^2} + 1}{\sqrt[4]{x}} \, dx. $$
Explanation of Method
Integrands built from roots of $x$ are easiest to handle by rewriting every root as a rational power of $x$. Once the integrand is a sum of terms of the form $x^n$, each term can be integrated directly with the power rule $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ (valid for $n \neq -1$).
- Rewrite every radical as $x$ raised to a rational exponent.
- Divide each term in the numerator by the denominator, subtracting exponents.
- Integrate each resulting power of $x$ term by term.
- Combine the results and add the constant of integration.
Step 1 · Rewrite the terms with rational exponents
The numerator of the integrand is:
$$ \sqrt{x} - 2\sqrt[3]{x^2} + 1, $$
which can be rewritten as:
$$ x^{1/2} - 2x^{2/3} + 1. $$
The denominator is:
$$ \sqrt[4]{x} = x^{1/4}. $$
Thus, the integrand becomes:
$$ \frac{\sqrt{x} - 2\sqrt[3]{x^2} + 1}{\sqrt[4]{x}} = \frac{x^{1/2}}{x^{1/4}} - \frac{2x^{2/3}}{x^{1/4}} + \frac{1}{x^{1/4}}. $$
Simplify each term by subtracting exponents:
$$ \frac{x^{1/2}}{x^{1/4}} = x^{1/2 - 1/4} = x^{1/4}, \quad \frac{2x^{2/3}}{x^{1/4}} = 2x^{2/3 - 1/4} = 2x^{5/12}, \quad \frac{1}{x^{1/4}} = x^{-1/4}. $$
So the integral becomes:
$$ I = \int \left( x^{1/4} - 2x^{5/12} + x^{-1/4} \right) dx. $$
Step 2 · Integrate each term
The integral splits into three simpler terms:
$$ I = \int x^{1/4}\,dx - 2\int x^{5/12}\,dx + \int x^{-1/4}\,dx. $$
First term: $\displaystyle \int x^{1/4}\,dx$. Using the power rule:
$$ \int x^{1/4}\,dx = \frac{x^{1/4+1}}{1/4+1} = \frac{x^{5/4}}{5/4} = \frac{4}{5}x^{5/4}. $$
Second term: $\displaystyle -2\int x^{5/12}\,dx$. Using the power rule:
$$ \int x^{5/12}\,dx = \frac{x^{5/12+1}}{5/12+1} = \frac{x^{17/12}}{17/12} = \frac{12}{17}x^{17/12}. $$
Multiplying by $-2$:
$$ -2\int x^{5/12}\,dx = -2 \cdot \frac{12}{17}x^{17/12} = -\frac{24}{17}x^{17/12}. $$
Third term: $\displaystyle \int x^{-1/4}\,dx$. Using the power rule:
$$ \int x^{-1/4}\,dx = \frac{x^{-1/4+1}}{-1/4+1} = \frac{x^{3/4}}{3/4} = \frac{4}{3}x^{3/4}. $$
Step 3 · Combine the results
Combining all three terms:
$$ I = \frac{4}{5}x^{5/4} - \frac{24}{17}x^{17/12} + \frac{4}{3}x^{3/4} + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Rewritten as | Antiderivative |
|---|---|---|
| $\sqrt{x} / \sqrt[4]{x}$ | $x^{1/4}$ | $\dfrac{4}{5}x^{5/4}$ |
| $2\sqrt[3]{x^2} / \sqrt[4]{x}$ | $2x^{5/12}$ | $\dfrac{24}{17}x^{17/12}$ |
| $1 / \sqrt[4]{x}$ | $x^{-1/4}$ | $\dfrac{4}{3}x^{3/4}$ |
Additional Notes
- The technique of converting radicals to rational exponents before dividing term-by-term works whenever a sum of root expressions is divided by a single radical.
- The domain of $I$ requires $x > 0$, since fourth and cube roots of $x$ combined with fractional exponents are defined for positive $x$ in the real-valued sense used here.
- The result can be checked by differentiating $\dfrac{4}{5}x^{5/4} - \dfrac{24}{17}x^{17/12} + \dfrac{4}{3}x^{3/4}$ and confirming it returns the original integrand.
Result
The solution to the integral is: