Question

高数求不定积分

第九题

Answer 1

原积分
=∫dx/(9x²-6x+2)
=∫dx/[(3x-1)²+1]
=1/3*∫d(3x-1)/[(3x-1)²+1]
=1/3*arctan(3x-1) +C
其中C为常数

Answer 2

(9)
9x^2-6x+2
=9[x^2- (2/3)x + 1/9] +1
=9( x-1/3)^2 +1
let
x-1/3 = (1/3)tanu
dx= (1/3)(secu)^2 du
∫ dx/(9x^2-6x+2)
=∫ (1/3)(secu)^2 du/ (secu)^2
=(1/3)u + C
=(1/3)arctan(3x-1) + C