Sec 6.3

1172 days 전, jhlee2chn 작성

보기 3.

f(x)=(x^4-2*x^2+4*x+1)/(x^3-x^2-x+1) integral(f(x), x) 
       
1/2*x^2 + x - 2/(x - 1) - log(x + 1) + log(x - 1)
1/2*x^2 + x - 2/(x - 1) - log(x + 1) + log(x - 1)
fp=f(x).partial_fraction() fp 
       
x - 1/(x + 1) + 1/(x - 1) + 2/(x - 1)^2 + 1
x - 1/(x + 1) + 1/(x - 1) + 2/(x - 1)^2 + 1
integral(fp, x) 
       
1/2*x^2 + x - 2/(x - 1) - log(x + 1) + log(x - 1)
1/2*x^2 + x - 2/(x - 1) - log(x + 1) + log(x - 1)

보기 4.

f(x)=(2*x^2-x+4)/(x^3+4*x) fp=f(x).partial_fraction() # 부분 분수 찾는 법 fp 
       
(x - 1)/(x^2 + 4) + 1/x
(x - 1)/(x^2 + 4) + 1/x

보기 5.

f(x)=(4*x^2-3*x+2)/(4*x^2-4*x+3) fp=f(x).partial_fraction() # 부분 분수 찾는 법 fp 
       
(x - 1)/(4*x^2 - 4*x + 3) + 1
(x - 1)/(4*x^2 - 4*x + 3) + 1
h(x)=4*x^2-3*x+2 k(x)=4*x^2-4*x+3 h.maxima_methods().divide(k) 
       
[1, x - 1]
[1, x - 1]

보기 7.

f=(1-x+2*x^2-x^3)/(x*(x^2+1)^2) fp=f.partial_fraction(x) fp 
       
-(x + 1)/(x^2 + 1) + 1/x + x/(x^2 + 1)^2
-(x + 1)/(x^2 + 1) + 1/x + x/(x^2 + 1)^2