用分部积分法求这两个不定积分

Question

如题所述

Answer 1

如下

追答

本回答被提问者采纳

Answer 2

(1)

u=√x
2u du = dx
∫ arctan√x/√x dx
=∫ [arctanu/u ] ( 2u du)
=2∫ arctanu du
=2u.arctanu-2∫ u/(1+u^2) du
=2u.arctanu-2ln|1+u^2| + C
=2√x.arctan√x-2ln|1+x| + C
(2)
x^2 = x(x+1) -x = x(x+1) -(x+1) +1
x^2= x(x-1) +x = x(x-1) +(x-1) +1
∫ x[ ln(x+1)/(1-x) ] dx
=∫ x [ln(x+1) -ln(1-x)] dx
=(1/2)∫ [ln(x+1) -ln(1-x)] dx^2
=(1/2)x^2.[ln(x+1) -ln(1-x)] - (1/2)∫ [ x^2/(x+1) -x^2/(1-x)] dx
=(1/2)x^2.[ln(x+1) -ln(1-x)] - (1/2)∫ [ x^2/(x+1) +x^2/(x-1)] dx
=(1/2)x^2.[ln(x+1) -ln(1-x)] - (1/2)∫{ [ x-1 +1/(x+1)] +[ x+1+ 1/(x-1)] } dx
=(1/2)x^2.[ln(x+1) -ln(1-x)] - (1/2)∫ [ 2x +1/(x+1)+ 1/(x-1)] dx
=(1/2)x^2.[ln(x+1) -ln(1-x)] - (1/2)[ x^2 +ln|x^2-1| ] + C本回答被网友采纳