∫ dx/(x² + 2x + 3)
= ∫ dx/[(x + 1)² + 2]
令z = (x + 1)/√2,dz = (1/√2)dx
= ∫ √2•dz/(2z² + 2)
= √2/2•∫ dz/(z² + 1)
= (1/√2)arctan(z) + C
= (1/√2)arctan[(x + 1)/√2] + C有點(diǎn)不太明白，我剛剛學(xué)不定積分。。令z = (x + 1)/√2，dz = (1/√2)dx這思路是怎樣的啊？z = (x + 1)/√2 = x/√2 + 1/√2dz = (x/√2)'dx + (1/√2)'dx = (1/√2)dx這個(gè)會吧？這題要運(yùn)用公式∫ dy/(y² + a²) = (1/a)arctan(y/a)要令(x + 1)² + 2變?yōu)閤² + a²，所以設(shè)√2*z = (x + 1)，兩邊平方后，你會發(fā)現(xiàn)2z = (x + 1)²，代入得2z² + 2 = 2(z² + 1)，就是x² + a²的形式，這里y = z，a = 1