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### Section 5.2 : Computing Indefinite Integrals

8. Evaluate $$\displaystyle \int{{\frac{4}{{{x^2}}} + 2 - \frac{1}{{8{x^3}}}\,dx}}$$.

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Hint : Don’t forget to move the $$x$$’s in the denominator to the numerator with negative exponents.
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We first need to move the $$x$$’s in the denominator to the numerator with negative exponents.

$\int{{\frac{4}{{{x^2}}} + 2 - \frac{1}{{8{x^3}}}\,dx}} = \int{{4{x^{ - 2}} + 2 - \frac{1}{8}{x^{ - 3}}\,dx}}$

Remember that the “8” in the denominator of the third term stays in the denominator and does not move up with the $$x$$.

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Once we’ve gotten the $$x$$’s out of the denominator there really isn’t too much to do other than to evaluate the integral.

\begin{align*}\int{{\frac{4}{{{x^2}}} + 2 - \frac{1}{{8{x^3}}}\,dx}} & = \int{{4{x^{ - 2}} + 2 - \frac{1}{8}{x^{ - 3}}\,dx}}\\ & = 4\left( {\frac{1}{{ - 1}}} \right){x^{ - 1}} + 2x - \frac{1}{8}\left( {\frac{1}{{ - 2}}} \right){x^{ - 2}} + c = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 4{x^{ - 1}} + 2x + \frac{1}{{16}}{x^{ - 2}} + c}}\end{align*}

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.