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

1 Answer
Feb 17, 2017

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

Explanation:

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