How do you evaluate #int (x^2)/(x^3+27)dx#?

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Answer 1 · 2018-05-20T14:14:47

The answer is #=1/3ln(|x^3+27|)+C#

Perform this integral by substitution,

Let #u=x^3+27#

#=>#, #du=3x^2dx#

The integral is

#int(x^2dx)/(x^3+27)=1/3int(du)/(u)#

#=1/3lnu#

#=1/3ln(|x^3+27|)+C#

Answer 2 · 2018-05-20T14:30:02

#ln(abs[x^3+27])/3+c#

Problem:
#intx^2/(x^3+27)dx#

Substitute #u=x^3+27->dx=1/(3x^2)du#
#=1/3int1/udu#

Now solving #int1/udu#

This is a standard integral
#=lnu#

Plug in solved integrals:
#1/3int1/udu#
=#ln(u)/3#

Undo substitution #u=x^3+27#
=#ln(x^3+27)/3#

Add a constant lastly and absolute #x^3+27# because you cannot #ln# a negative number.