0
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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

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


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

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

$\endgroup$
1
$\begingroup$

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

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

By the reverse power rule.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.