求不定积分∫1/√(5-4x+x^2) dx

∫1/√(5-4x+x^2) dx
=∫1/√((x-2)^2+1) dx
令x-2=tant
则原式=∫1/√(（tant)^2+1)d(tant+2)
=∫1/√（1/(cost)^2) dtant
=∫1/√（1/(cost)^2) dtant
=∫costdtant
=∫cost•(1/(cost)^2)dt
=∫cos/(1-(sint)^2) dt
=∫dsint/(1-(sint)^2)
=(1/2)∫ (1/(1+sint)+1/(1-sint))d(sint)
=(1/2) (ln|1+sint|-ln|1-sint| )+ C
=ln √((1+sint)/(1-sint)) + C
=ln √(1+sint)²/√(1-sin²t) + C
=ln |(1+sint)/cost| + C
=ln |tant+sect| + C
=ln|x-2+√((x-2)²+1)|+C