Question

如图不定积分怎么算？

Answer 1

x^5+2x^3+x
=x(x^4+2x^2 +1)
=x(x^2+1)^2

let
(3x^4+x^3+4x^2 +1)/(x^5+2x^3+x)
≡A/x+ (Bx+C)/(x^2+1) + (Dx+E)/(x^2+1)^2

=>
3x^4+x^3+4x^2 +1≡A(x^2+1)^2+ (Bx+C)x(x^2+1) + (Dx+E)x

x=0 => A=1

coef. of x^4
A+B= 3
B=2

coef. of x^3, =>C=1

coef. of x
C+E =0
E =-1

x=1

9=4A+ 2(B+C) + (D+E)
9=4+2(2+1)+(D-1)
9= 9+D
D=0

(3x^4+x^3+4x^2 +1)/(x^5+2x^3+x)
≡1/x+ (2x+1)/(x^2+1) - 1/(x^2+1)^2

∫(3x^4+x^3+4x^2 +1)/(x^5+2x^3+x) dx
=∫ [1/x+ (2x+1)/(x^2+1) - 1/(x^2+1)^2 ] dx
=∫ (1/x) dx + ∫2x/(x^2+1) dx +∫ dx/(x^2+1) - ∫ dx/(x^2+1)^2
=ln|x| +ln|x^2+1| + arctanx -∫ dx/(x^2+1)^2
=ln|x| +ln|x^2+1| + arctanx -(1/2) [ arctanu + x/(1+x^2) ] + C
=ln|x| +ln|x^2+1| + (1/2)arctanx -(1/2)[x/(1+x^2) ] + C

let
x= tanu
dx = (secu)^2 .du
∫ dx/(x^2+1)^2
=∫ (cosu)^2 du
=(1/2) ∫ (1+ cos2u) du
=(1/2) [ u + (1/2)sin2u] + C'
=(1/2) [ arctanu + x/(1+x^2) ] + C'