Question

求不定积分。谢谢~

1.∫x/(3-2x^2) du
2.∫dx/√x +1

谢谢 需要过程~~

Answer 1

∫ x/(3 - 2x²) dx
= ∫ 1/(3 - 2x²) d(x²/2)
= (1/2)(-1/2)∫ 1/(3 - 2x²) d(-2x²)
= (-1/4)∫ 1/(3 - 2x²) d(3 -2x²)
= (-1/4)ln|3 - 2x²| + C

∫ dx/(√x + 1)
= ∫ 1/(√x + 1) * (2√x)/(2√x) dx
= 2∫ (√x + 1 - 1)/(√x + 1) * d√x
= 2∫ d√x - 2∫ 1/(√x + 1) d(√x + 1)
= 2√x - 2ln(1 + √x) + C

两题都是凑微分法