Question

高等数学不定积分求解！

Answer 1

x^2+2x+2 = (x+1)^2 +1,
令 x+1 = tanu, 则 dx = (secu)^2du
I = ∫ (secu)^2du /(1+secu)
= ∫ [(secu)^2-1+1]du /(1+secu)
= ∫ [secu-1+1/(1+secu)]du
= ln|secu+tanu| - u + ∫cosudu/(1+cosu)
= ln|secu+tanu| - u
+ ∫{[cos(u/2)]^2 - [sin(u/2)]^2}du/{2[cos(u/2)]^2}
= ln|secu+tanu| - u
+ ∫{[cos(u/2)]^2 - [sin(u/2)]^2}d(u/2)/[cos(u/2)]^2
= ln|secu+tanu| - u + ∫{1 - [tan(u/2)]^2}d(u/2)
= ln|secu+tanu| - u + ∫{2 - [sec(u/2)]^2}d(u/2)
= ln|secu+tanu| + tan(u/2) + C
= ln|√(x^2+2x+2)+x+1| + [√(x^2+2x+2)-1]/(x+1) + C