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How do you use partial fractions to find the integral ?

#int (dx)/(x^3+8#

Answer 1

Elijah Abram

The answer is #=1/12ln(|x+2|)-1/24ln(|x^2-2x+4|)+sqrt3/12arctan((x-1)/sqrt3)+C#

#x^3+8=(x+2)(x^2-2x+4)#

Perform the decomposition into partial fractions

#1/(x^3+8)=1/((x+2)(x^2-2x+4))#

#=A/((x+2))+(Bx+C)/(x^2-2x+4)#

#=(A(x^2-2x+4)+(Bx+C)(x+2))/((x+2)(x^2-2x+4))#

The denominators are the same, compare the numerators

#1=A(x^2-2x+4)+(Bx+C)(x+2)#

Let #x=-2#, #=>#, #1=12A#, #=>#, #A=1/12#

Let #x=0#, #=>#, #1=4A+2C#, #=>#, #C=1/3#

Coefficients of #x^2#

#0=A+B#, #=>#, #B=-1/12#

Therefore,

#1/(x^3+8)=(1/12)/((x+2))+(-1/12x+4/12)/(x^2-2x+4)#

#int(1dx)/(x^3+8)=int(1/12dx)/((x+2))+int((-1/12x+4/12)dx)/(x^2-2x+4)#

#=1/12ln(|x+2|)-1/12int((x-4)dx)/(x^2-2x+4)#

Let #x-4=1/2(2x-2)-3#

Perform the second integral separately

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

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

#x^2-2x+4=x^2-2x+1+3=(x-1)^2+(sqrt3)^2#

#=sqrt3(((x-1)/sqrt3)^2+1)#

So,

#3int(dx)/(x^2-2x+4)=3/sqrt3int(dx)/(((x-1)/sqrt3)^2+1)#

#=sqrt3arctan((x-1)/sqrt3)#

Finally,

#int(1dx)/(x^3+8)=1/12ln(|x+2|)-1/24ln(|x^2-2x+4|)+sqrt3/12arctan((x-1)/sqrt3)+C#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.