Section 5.2 : Computing Indefinite Integrals
13. Evaluate \( \displaystyle \int{{\frac{{{x^4} - \sqrt[3]{x}}}{{6\sqrt x }}\,dx}}\).
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Hint : Remember that there is no “Quotient Rule” for integrals and so we’ll need to eliminate the quotient before integrating.
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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!
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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.