Question

一道数学不定积分小题 求详解

Answer 1

求不定积分∫[(x+2)/√(4x²-4x+5)]dx
解：原式=∫{(x+2)/√[4(x²-x)+5]}dx=(1/2)∫(x+2)/√[(x-1/2)²-1/4+5/4]dx=(1/2)∫(x+2)/√[(x-1/2)²+1]}dx
【令x-1/2=tanu，则x=(1/2)+tanu，dx=sec²udu，代入上式得】
=(1/2)∫[(tanu+(5/2)]sec²udu/√(1+tan²u)=(1/2)∫[(tanu+(5/2)]secudu
=(1/2)∫sinusec²udu+(5/4)∫secdu=(1/2)∫sinud(tanu)+(5/4)ln(secu+tanu)
=(1/2)[sinutanu-∫tanucosudu]+(5/4)ln(secu+tanu)
=(1/2)[sinutanu-∫sinudu]+(5/4)ln(secu+tanu)
=(1/2)(sinutanu+cosu)+(5/4)ln(secu+tanu)+C
=(1/2){(2x-1)²/[2√(4x²-4x+5)+2/√(4x²-4x+5)]+(5/4)ln[2/√(4x²-4x+5)+(2x-1)/2]+C
=(4x²-4x+5)/[4√(4x²-4x+5)+(5/4)ln[2/√(4x²-4x+5)+(2x-1)/2]+C
=(1/4)√(4x²-4x+5)+(5/4)ln[2/√(4x²-4x+5)+(2x-1)/2]+C

Answer 2

解：
∫(2+x)/(x^2-4x+5) dx
=∫(x-2+4)/((x-2)^2+1))dx
=∫(x-2)/((x-2)^2+1)dx+4∫4/((x-2)^2+1)dx
=lnl(x-2)^2+1)l + 4arctan(x-2)
=ln lx^2-4x+5l + 4arctan(x-2) + C