首先你要懂得導(dǎo)數(shù)的運算公式,求不定積分是求導(dǎo)的逆過程∫ x/(1 + x²) dx= ∫ 1/(1 + x²) • (x dx)= ∫ 1/(1 + x²) d(x²/2)這里其實是對x求積分的,即x dx ∫ x dx = x²/2 + C d(x...你看看我的分析對不dx^2=（x^2）’dx分析df(x)/dx=f(x)'一樣只不過f(x)換成了x^2.如果我分析對的話du=d(u+1)我也明白了。對d(x²) = (x²)' dx = (2x) dx,，兩邊除以2得==> (1/2) d(x²) = x dx 或 x dx = d(x²/2)du = d(u ± 1) = d(u ± 1000) = d(u ± 任意常數(shù))但du = d(Ku/K) = (1/K) d(Ku) = (1/K) d(Ku + N)或du = d(u/K • K) = K d(u/K) = K d(u/K + M)加減常數(shù)可任意加上，但乘以常數(shù)需要抵消 !