∫(x+2)/√(2x+1)dx,积分限为(0,4)
令2x+1 = t,x = (t-1)/2
积分限变为(1,9)
∫(x+2)/√(2x+1)dx,积分限为(0,4)
= ∫[(t-1)/2+2]/√t d(t-1)/2,积分限为(1,9)
= ∫(t+3)/√t dt
= ∫[√t/4+3/(4√t)] dt
= {t^(3/2)/6 + 3√t/2},t=9 时减去t=1时
= 9 - 5/3
= 22/3
∫4 0(x+2)/根号(2x+1)求定积分
∫(x+2)/√(2x+1)dx,积分限为(0,4)
令2x+1 = t,x = (t-1)/2
积分限变为(1,9)
∫(x+2)/√(2x+1)dx,积分限为(0,4)
= ∫[(t-1)/2+2]/√t d(t-1)/2,积分限为(1,9)
= ∫(t+3)/√t dt
= ∫[√t/4+3/(4√t)] dt
= {t^(3/2)/6 + 3√t/2},t=9 时减去t=1时
= 9 - 5/3
= 22/3