用换元法求不定积分

如图

1. 令 √(2x) = u, 则 x = u^2/2, dx = udu
I = ∫ udu/(u-1) = ∫ [1+1/(u-1)]du
= u + ln|u-1| + C = √(2x) + ln|√(2x)-1| + C
2. 令 √(1+e^x) = u, 则 e^x = u^2-1, x = ln(u^2-1), dx = 2udu/(u^2-1),
I = ∫ 2du/(u^2-1) = ∫ [1/(u-1) - 1/(u+1)]du
= ln|(u-1)/(u+1)| + C = ln|[√(1+e^x)-1]/[√(1+e^x)+1]| + C
= 2ln|√(1+e^x)-1| - x + C
5. 令 x = tanu, 则 dx = (secu)^2 du,
I = ∫ (tanu)^3(secu)^3du = ∫ (sinu)^3du/(cosu)^6
= ∫ [(cosu)^2-1]dcosu/(cosu)^6 = ∫ [(cosu)^(-4) - (cosu)^(-6)]dcosu
= (-1/3)(cosu)^(-3) + (1/5)(cosu)^(-5) + C
= (-1/3)/(cosu)^3 + (1/5)/(cosu)^5 + C
= (-1/3)(1+x^2)^(3/2) + (1/5)/(1+x^2)^(5/2) + C
6. 令 x = sinu, 则 dx = cosudu,
I = ∫ (cosu)^2du/(sinu)^4 = - ∫ (cotu)^2dcotu = -(1/3)(cot)^3 + C
= -(1/3)(1-x^2)^(3/2)/x^3 + C