The integration of ?#int((x^2-1)/(x^3-3x))# dx

Question

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

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Answer 1 · 2018-04-13T14:38:28

The answer is #=1/3ln(|x^3-3x|)+C#

Perform this integral by substitution

Let #u=x^3-3x#

#du=3x^2-3=3(x^2-1)dx#

Therefore, the integral is

#int((x^2-1)dx)/(x^3-3x)=1/3int(du)*1/u#

#=1/3lnu#

#=1/3ln(|x^3-3x|)+C#

Answer 2 · 2018-04-13T14:42:08

# 1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C#.

Suppose that, #I=int(x^2-1)/(x^3-3x)dx#.

Observe that, #d/dx(x^3-3x)=3x^2-3=3(x^2-1)#.

Knowing that, #int(f'(x))/f(x)dx=ln|f(x)|+c#, we have,

#I=1/3int(3(x^2-1))/(x^3-3x)dx#.

# rArr I=1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C#.