Question

不定积分求解答

Answer 1

x^3
= x(x^2-2x+2) + 2x^2-2x
= x(x^2-2x+2) + 2(x^2-2x+2) +2x-4
∫x^3/(x^2-2x+2)^2 dx
=∫(x+2)/(x^2-2x+2) dx +∫ (2x-4)/(x^2-2x+2)^2 dx
=(1/2)∫(2x-2)/(x^2-2x+2) dx +3∫dx/(x^2-2x+2)
+∫ (2x-2)/(x^2-2x+2)^2 dx -2∫ dx/(x^2-2x+2)^2
=(1/2)ln|x^2-2x+2| +3∫dx/(x^2-2x+2) -[1/(x^2-2x+2)] -2∫ dx/(x^2-2x+2)^2
=(1/2)ln|x^2-2x+2|+3arctan(x-1) -[1/(x^2-2x+2)] -[arctan(x-1)+(x-1)/(x^2-2x+2)] + C
=(1/2)ln|x^2-2x+2|+2arctan(x-1) -[1/(x^2-2x+2)] - [(x-1)/(x^2-2x+2)] + C
/
∫dx/(x^2-2x+2)
=∫dx/[(x-1)^2+1]
=∫d(x-1)/[(x-1)^2+1]
=arctan(x-1) + C1
/
let
x-1 = tanu
dx = (secu)^2 du
∫ dx/(x^2-2x+2)^2
=∫ dx/[(x-1)^2+1]^2
=∫ (secu)^2 du/(secu)^4
=∫ (cosu)^2 du
=(1/2)∫ (1+cos2u) du
=(1/2)[ u+(1/2)sin2u] + C2
=(1/2)[ arctan(x-1)+ (x-1)/(x^2-2x+2) ] + C2