How do you integrate #int 1/(x^3 +4x) dx# using partial fractions?

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Answer 1 · 2017-02-17T13:41:32

#int dx/(x^3+4x) = 1/8ln (x^2 / (x^2+4) )+C#

Factorize the denominator:

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

So:

#1/(x^3+4x) = A/x + (Bx+C)/(x^2+4)#

#1/(x^3+4x) = (A(x^2+4) + x(Bx+C))/(x(x^2+4))#

#1/(x^3+4x) = (Ax^2+4A + Bx^2+Cx)/(x(x^2+4))#

#1/(x^3+4x) = ((A+B)x^2 + Cx + 4A)/(x(x^2+4))#

and equating the coefficient with the same degree in #x#:

#{(A+B=0),(C=0),(4A=1):}#

#{(A=1/4),(B=-1/4),(C=0):}#

We have then:

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

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

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

that we can also write as:

#int dx/(x^3+4x) = 1/8ln (x^2 / (x^2+4) )+C#