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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.
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 \(x\).
Show Step 2Once 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.