How do you integrate #int (2x) / (x^3 + 1)# using partial fractions?

1 Answer
Roy E.
Dec 21, 2016

#-2/3ln|x+1|+1/3ln|x^2-x+1|+2/sqrt3tan^-1((2x-1)/sqrt3)+C#

Explanation:

The denominator factorises as #(x+1)(x^2-x+1)#. So the integrand is
#(-2/3)/(x+1)+(Ax+B)/(x^2-x+1)# where the #2/3# comes from the cover-up rule: #((2)(-1))/((-1)^2-(-1)+1)# and #A# and #B# are to be found by equating coefficients of suitable powers of #x# or otherwise:
#2x-=(-2/3)(x^2-x+1)+(Ax+B)(x+1)# which, upon re-grouping in powers of #x# gives:
#0x^2+2x+0-=(-2/3+A)x^2+(2/3+A+B)x-2/3+B#.
So #A=2/3#, #B=2/3#

So the integral is:
#int 2/3(-1/(x+1)+1/2(2x-1+3)/(x^2-x+1))dx#
#=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+int1/((x-1/2)^2+3/4)dx#
#=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+(2/sqrt3)tan^-1((x-1/2)/(sqrt3/2))+C#
#=2/3(-ln|x+1|)+1/3ln|x^2-x+1|+(2/sqrt3)tan^-1((2x-1)/sqrt3)+C#

Notice the trick of splitting the numerator #x+1# into the multiple of the derivative of the denominator plus a correction constant (here #+3#) in anticipation of using the rule about integrating #int(f'(x))/f(x)dx=ln|f(x)|+C#

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