Evaluate the indefinite integral of x^2 - 13*x^(2/4) with respect to x.

Step-by-step solution

1. 1

   Reduce the exponent to lowest terms. $\tfrac{2}{4} = \tfrac12$, so $x^{2/4} = x^{1/2} = \sqrt{x}$, and the integrand is

   $x^2 - 13\sqrt{x}$

   Leaving the exponent unreduced would not be wrong, but it makes the $n+1$ arithmetic in the next step needlessly awkward.

2. 2

   Split the integral by linearity.

   $\int \left(x^2 - 13x^{1/2}\right)dx = \int x^2\,dx - 13\int x^{1/2}\,dx$

3. 3

   Apply the power rule to each term. With $\int x^n dx = \tfrac{x^{n+1}}{n+1} + C$ (valid since neither exponent is $-1$):

   $\int x^2\,dx = \frac{x^3}{3}, \qquad \int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$

4. 4

   Multiply through by the coefficient.

   $-13 \cdot \frac{2}{3}x^{3/2} = -\frac{26}{3}x^{3/2}$

5. 5

   Assemble the answer.

   $\int \left(x^2 - 13\sqrt{x}\right)dx = \frac{x^3}{3} - \frac{26}{3}x^{3/2} + C$

6. 6

   Differentiate to check. $\tfrac{d}{dx}\left(\tfrac{x^3}{3}\right) = x^2$ and $\tfrac{d}{dx}\left(-\tfrac{26}{3}x^{3/2}\right) = -\tfrac{26}{3}\cdot\tfrac32 x^{1/2} = -13\sqrt{x}$. The derivative returns the integrand exactly. Note the domain: $\sqrt{x}$ requires $x \ge 0$.