I'm trying to understand how to solve this indefinite integral using the power rule.

$$ \begin{eqnarray} \int \frac{x}{\sqrt{x}} dx &=& \int \frac{x^1}{x^\frac{1}{2}} dx \end{eqnarray} $$

Breaking this down

$$ \int x dx = \frac{x^2}{2} + C $$

$$ \begin{eqnarray} \int x^\frac{1}{2} dx &=& \frac{x^\frac{3}{2}}{\frac{3}{2}} + C\\ &=& \frac{2}{3} x^\frac{3}{2} + C \end{eqnarray} $$

Wolfram Alpha gives the integral of the original denominator ($\int\sqrt{x} dx$) as the solution. I do not understand what happened to the numerator?

How do I bring the two parts together?

  • 2
    $\begingroup$ $$\frac {x}{x^{1/2}} = x^{1-\frac 12} = x^{1/2}$$ So you are looking for $$\int x^{1/2} dx = \frac{x^{3/2}}{3/2} + C = \frac 23 \cdot x^{3/2} +C$$ $\endgroup$ – Namaste Mar 11 '18 at 16:01
  • $\begingroup$ There is no simple rule for integrating a product or a quotient. You have to simplify your algebra. What is $\dfrac{x}{\sqrt x}$? $\endgroup$ – Ted Shifrin Mar 11 '18 at 16:01
  • $\begingroup$ ${x\over \sqrt x}= \sqrt x$? $\endgroup$ – Andrew Li Mar 11 '18 at 16:02
  • $\begingroup$ The "power rule" would be $\frac{x^a}{x^b} = x^{a-b}$ as used by @amWhy's comment. $\endgroup$ – Jeppe Stig Nielsen Mar 11 '18 at 16:07
  • $\begingroup$ Thank you @amWhy and and Ted thats very clear $\endgroup$ – clicky Mar 11 '18 at 16:25

$$\frac{x}{\sqrt{x}} = \frac{\sqrt{x}\sqrt{x}}{\sqrt{x}} = \sqrt{x}$$

$$\int \sqrt{x}\ dx = \frac{2}{3}x^{3/2} + c$$

Note that $\sqrt{x} = x^{1/2}$ hence when you integrate it, just apply the integration rule for $x^a$.

$$\int x^a\ dx = \frac{x^{a+1}}{a+1} + c ~~~~~~~~~~~ \text{for} ~~~ a\neq -1$$


Hint: You can simplify $x\over \sqrt x$ into just $\sqrt x$. Then integrate:

$$\int \sqrt x \, dx$$

By the reverse power rule.


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