What is #int (-3x^3+4 ) / (-2x^2-2x +1 )#?

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Answer 1 · 2018-05-06T11:09:33

#I=(3x^2)/4-(3x)/2+9/8ln|x^2+x-1/2|-31/(8sqrt3)ln|(2x+1-sqrt3)/(2x+1+sqrt3)|+C#

Here,

#I=int(-3x^3+4)/(-2x^2-2x+1)dx#

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

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

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

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

#=3/2intxdx-3/2int(x^2-x/2+4/3)/(x^2+x-1/2)dx#

#=3/2*x^2/2-3/2int((x^2+x-1/2)-((3x)/2-11/6))/(x^2+x-1/2)dx#

#=(3x^2)/4-3/2int1dx+3/2int((3x)/2-11/6)/(x^2+x-1/2)dx#

#=(3x^2)/4-(3x)/2+3/2xx3/2int(x-11/9)/(x^2+x-1/2)dx#

#=(3x^2)/4-(3x)/2+9/4xx1/2int(2x-22/9)/(x^2+x-1/2)dx#

#=(3x^2)/4-(3x)/2+9/8int((2x+1)-31/9)/(x^2+x-1/2)dx#

#=(3x^2)/4-(3x)/2+9/8int(2x+1)/(x^2+x-1/2)dx- 9/8int(31/9)/(x^2+x-1/2)dx/#

#=(3x^2)/4-(3x)/2+9/8ln|x^2+x-1/2|-9/8xx31/9I_1...to(A)#

#Where, I_1=int1/(x^2+x+1/4-3/4)dx=int1/((x^+1/2)^2- (sqrt3/2))dx#

#:.I_1=1/(2(sqrt3/2))ln|(x+1/2-sqrt3/2)/(x+1/2+sqrt3/2)|+c#

#I_1=1/sqrt3ln|(2x+1-sqrt3)/(2x+1+sqrt3)|+c#

Hence, from #(A)#,

#I=(3x^2)/4-(3x)/2+9/8ln|x^2+x-1/2|-31/(8sqrt3)ln|(2x+1- sqrt3)/(2x+1+sqrt3)|+C#