Qual é a integral de #int 1 / (x ^ 3-1) dx #?

Responda:

#I=1/3ln|x-1|-1/6ln|x^2+x+1|-1/sqrt3tan^-1((2x+1)/sqrt3)+c#

Explicação:

Aqui,

#I=int1/(x^3-1)dx#

#=int1/((x-1)(x^2+x+1))dx#

Toma,

#M=1/((x-1)(x^2+x+1))=A/(x-1)+(Bx+C)/(x^2+x+1)#

#=>1=A(x^2+x+1)+(Bx+C)(x-1)#

#=>1=x^2(A+B)+x(A+B-C)+(A-C)#

Comparando coeficiente de #x^2,x# e termo constante de ambos os lados

#A+B=0to(1) , A+B-C=0to(2),A-C=1to(3)#

Resolvendo, temos #A=1/3, B=-1/3, C=-2/3#

Assim,

#I=int(1/3)/(x-1)dx+int(-1/3x-2/3)/(x^2+x+1)dx#

#=1/3ln|x-1|-1/3int(x+2)/(x^2+x+1)dx#

#=1/3ln|x-1|-1/6int(2x+4)/(x^2+x+1)dx#

#=1/3ln|x-1|-1/6int(2x+1+3)/(x^2+x+1)dx#

#=1/3ln|x-1|-1/6int(2x+1)/(x^2+x+1)dx-3/6int1/(x^2+x+1)dx#

#=1/3ln|x-1|-1/6ln|x^2+x+1|-1/2int1/(x^2+x+1/4+3/4)dx#

#=1/3ln|x-1|-1/6ln|x^2+x+1|-1/2intdx/((x+1/2)^2+(sqrt3/2)^2)#

#=1/3ln|x-
1|-1/6ln(x^2+x+1)-1/2xx1/(sqrt3/2)tan^-1((x+1/2)/(sqrt3/2))+c#

#=1/3ln|x-1|-1/6ln|x^2+x+1|-1/sqrt3tan^-1((2x+1)/sqrt3)+c#

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