Question

求下列定积分

Answer 1

1. I = ∫2cosxdx + ∫(lnx)^2d(lnx) - (1/2)∫e^(-x^2+1)d(-x^2+1)
= 2sinx + (1/3)(lnx)^3 - (1/2)e^(-x^2+1) + C;
2. 令 u = √(x+1)， 则 x+1 = u^2, dx = 2udu
I = ∫(u^2(2u)du/(u+1) = 2∫(u^3+u^2-u^2-u+u+1-1)du/(u+1)
= 2∫[u^2-u+1-1/(u+1)]du = (2/3)u^3 - u^2 + 2u - 2ln(u+1) + C1
= (2/3)(x+1)^(3/2) - (x+1) + 2√(x+1) - 2ln[√(x+1)+1] + C1
= (2/3)(x+1)^(3/2) - x + 2√(x+1) - 2ln[√(x+1)+1] + C
3. 令 u = lnx, 则 x = e^u,
∫<1, 4>sinlnxdx = ∫<0, ln4>sinu e^udu
I = ∫sinu e^udu = ∫sinu de^u = e^usinu - ∫e^ucosudu
= e^usinu - ∫cosude^u = e^usinu - e^ucosu - ∫e^usinudu
2I = e^u(sinu-cosu), I = (1/2)e^u(sinu-cosu)
∫<1, 4>sinlnxdx = ∫<0, ln4>sinu e^udu
= (1/2)[e^u(sinu-cosu)]<0, ln4> = (1/2)[4(sinln4-cosln4)+1]
= 2(sinln4-cosln4) + 1/2