Paul's Online Notes
Home / Calculus I / Integrals / Computing Indefinite Integrals
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 5.2 : Computing Indefinite Integrals

13. Evaluate $$\displaystyle \int{{\frac{{{x^4} - \sqrt[3]{x}}}{{6\sqrt x }}\,dx}}$$.

Show All Steps Hide All Steps

Hint : Remember that there is no “Quotient Rule” for integrals and so we’ll need to eliminate the quotient before integrating.
Start Solution

Since there is no “Quotient Rule” for integrals we’ll need to break up the integrand and simplify a little prior to integration.

$\int{{\frac{{{x^4} - \sqrt[3]{x}}}{{6\sqrt x }}\,dx}} = \int{{\frac{{{x^4}}}{{6{x^{\frac{1}{2}}}}} - \frac{{{x^{\frac{1}{3}}}}}{{6{x^{\frac{1}{2}}}}}\,dx}} = \int{{\frac{1}{6}{x^{\frac{7}{2}}} - \frac{1}{6}{x^{ - \,\frac{1}{6}}}\,dx}}$

Don’t forget to convert the roots to fractional exponents!

Show Step 2

At this point there really isn’t too much to do other than to evaluate the integral.

$\int{{\frac{{{x^4} - \sqrt[3]{x}}}{{6\sqrt x }}\,dx}} = \int{{\frac{1}{6}{x^{\frac{7}{2}}} - \frac{1}{6}{x^{ - \,\frac{1}{6}}}\,dx}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{27}}{x^{\frac{9}{2}}} - \frac{1}{5}{x^{\,\frac{5}{6}}} + c}}$

Don’t forget to add on the “+c” since we know that we are asking what function did we differentiate to get the integrand and the derivative of a constant is zero and so we do need to add that onto the answer.