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### Section 5.2 : Computing Indefinite Integrals

12. Evaluate $$\displaystyle \int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}}$$.

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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.
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.

$\int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}} = \int{{\frac{{{z^8}}}{{{z^4}}} - \frac{{6{z^5}}}{{{z^4}}} + \frac{{4{z^3}}}{{{z^4}}} - \frac{2}{{{z^4}}}\,dz}} = \int{{{z^4} - 6z + \frac{4}{z} - 2{z^{ - 4}}\,dz}}$ Show Step 2

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

$\int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}} = \int{{{z^4} - 6z + \frac{4}{z} - 2{z^{ - 4}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{5}{z^5} - 3{z^2} + 4\ln \left| z \right| + \frac{2}{3}{z^{ - 3}} + c}}$

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.