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How do you evaluate the integral #int dx/(x^3+1)#?

1 Answer
Jan 21, 2017

Answer:

Use partial fraction expansion and then integrate

Explanation:

Partial fraction expansion of #1/(x^3 +1)#:

-1 is a root of the denominator, therefore, #(x +1)# is a factor:

#color(white)((x +1)/color(black)(x+1)(color(black)((x^2 - x + 1)/("|"x^3 + 0x^2 + 0x + 1))#
#color(white)(" ")(-x^3 - x^2)/(" "-x^2#
#color(white)(" ")(+x^2 + x" ")/(" "x+1)#
#color(white)(" ")-x-1#

Check the discriminant of the quotient:

#b^2 - 4(a)(c) = -1^2 - 4(1)(1) = -3#

There are no real roots, therefore, the partial fractions are:

#1/(x^3 + 1) = (Ax + B)/(x^2 - x + 1) + C/(x + 1)#

Multiply both sides by #(x^2 -x + 1)(x + 1)#:

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

Make A and B disappear by letting x = -1:

#1 = C((-1)^2- -1 + 1)#

#C = 1/3#

#1 = (Ax + B)(x + 1) + 1/3(x^2 - x + 1)#

Make A disappear by letting x = 0:

#1 = B + 1/3#

#B = 2/3#

Let x = 1:

#1 = (A + 2/3)(2) + 1/3#
#1 = 2A + 5/3#
#A = -1/3#

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

Setting up the second term for "u" substitution:

#1/(x^3 + 1) = 1/3(1)/(x + 1) -1/6(2x - 4)/(x^2 - x + 1)#

We want #2x - 1# in the numerator of the second term, therefore we much create a third term for the remaining -3:

#1/(x^3 + 1) = 1/3(1)/(x + 1) -1/6(2x - 1)/(x^2 - x + 1) - 1/6(-3)/(x^2 - x + 1)#

Now, the first two terms will integrate to natural logarithms and the last term will be a complete the square integral to become the inverse tangent:

#1/(x^3 + 1) = 1/3(1)/(x + 1) -1/6(2x - 1)/(x^2 - x + 1) + 1/2(1)/(x^2 - x + 1)#

Write each term as an integral:

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

#int1/(x^3 + 1)dx = 1/3ln(x + 1) -1/6ln(x^2 - x + 1) + sqrt3/3tan^-1((2x - 1)/sqrt3) + C#