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### Section 5.2 : Computing Indefinite Integrals

17. Evaluate $$\displaystyle \int{{4{{\bf{e}}^z} + 15 - \frac{1}{{6z}}\,dz}}$$.

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There really isn’t too much to do other than to evaluate the integral.

$\int{{4{{\bf{e}}^z} + 15 - \frac{1}{{6z}}\,dz}} = \int{{4{{\bf{e}}^z} + 15 - \frac{1}{6}\frac{1}{z}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{4{{\bf{e}}^z} + 15z - \frac{1}{6}\ln \left| z \right| + c}}$

Be careful with the “6” in the denominator of the third term. The “best” way of dealing with it in this case is to split up the third term as we’ve done above and then integrate.

Note that the “best” way to do a problem is always relative for many Calculus problems. There are other ways of dealing with this term (later section material) and so what one person finds the best another may not. For us, this seems to be an easy way to deal with the 6 and not overly complicate the integration process.