Question

两道高数题

求解

Answer 1

求采纳

追答

第二题换元

Answer 2

∫(1→e) x²lnx dx
= ∫(1→e) lnx d(x³/3)
= (1/3)x³lnx |(1→e) - (1/3)∫(1→e) x³ d(lnx),分部积分法
= (1/3)(e³ - 0) - (1/3)∫(1→e) x³ * 1/x dx
= (1/3)e³ - (1/3) * x³/3 |(1→e)
= (1/3)e³ - (1/9)(e³ - 1)
= (2e³ + 1)/9
∫(0→5)1/√(1+3x)dx
令 t = √(1+3x) , 则 x = (t^2-1)/3 ， dx = 2t/3 dt
当 x = 0 时，t = 1；当 x=5 时，t = 4
原式 = ∫(1→4) 1/t * 2t/3 dt
=∫(1→4) 2/3 dt
= [2t/3](1→4)
= 2

追问

谢谢