PART 17：例題-避重就輕選擇

求不定積分 \(\int {{x^2}\sqrt {x + 3} \;dx} \) 我們希望根號內愈簡單愈好，所以令 \(u = x + 3\;\;\)  

\( \Rightarrow \;\;x = u - 3\;\;\) \(\; \Rightarrow \;\;dx = du\)

SOL:

原式=\(\int {{x^2}\sqrt {x + 3} \;dx} \)  

\( = \int {{{(u - 3)}^2}\sqrt u \;du} \) \( = \int {({u^2} - 6u + 9){u^{1/2}}\;du} \)  

\( = \int {({u^{5/2}} - 6{u^{3/2}} + 9{u^{1/2}})\;du} \)  

\( = \frac{2}{7}{u^{7/2}} - 6 \cdot \frac{2}{5}{u^{5/2}} + 9 \cdot \frac{2}{3}{u^{3/2}} + C\)  

\( = \frac{2}{7}{(x + 3)^{7/2}} - \frac{{12}}{5}{(x + 3)^{5/2}} + 6{(x + 3)^{3/2}} + C\)