Question

solve the following integral. the answer can be determined by rules of antidifferentiation, but some algebra may be required beforehand
int \frac { u ^ { 6 } - 3 u ^ { 4 } + 9 } { u ^ { 4 } } d u
int \frac { u ^ { 6 } - 3 u ^ { 4 } + 9 } { u ^ { 4 } } d u =

Explanation:

Step1: Simplify the integrand

Divide each term in the numerator by \(u^4\):

$$
LATEXBLOCK0
$$

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)):

• For \(\int u^{2}du\), by the power rule \(\int u^{2}du=\frac{u^{2+1}}{2 + 1}=\frac{u^{3}}{3}\)
• For \(\int(-3)du\), since \(\int kdu=ku + C\) (\(k=-3\)), we have \(\int(-3)du=-3u\)
• For \(\int9u^{-4}du\), using the power rule \(\int9u^{-4}du=9\times\frac{u^{-4 + 1}}{-4+1}=9\times\frac{u^{-3}}{-3}=- 3u^{-3}\)

Answer:

\(\frac{u^{3}}{3}-3u-\frac{3}{u^{3}}+C\)