PART 15：例題-遷就分母

求不定積分 \(\int {\frac{{{x^2}}}{{x + 2}}\;dx} \)

SOL:

(令 \(x + 2 = u\) )

\(\int {\frac{{{x^2}}}{{x + 2}}\;dx}  = \int {\frac{{{{(u - 2)}^2}}}{u}\;du} \)

\( = \int {\frac{{{u^2} - 4u + 4}}{u}\;du} \)  \( = \int {\left( {u - 4 + \frac{4}{u}\;} \right)du} \)

\( = \frac{1}{2}{u^2} - 4u + 4\ln u + C\)  

\( = \frac{1}{2}{(x + 2)^2} - 4(x + 2) + 4\ln (x + 2) + C\)    

註: \(x + 2 > 0\)