How to calculate #int(3x^2)/(x^2-2x-8)dx# ?

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Answer 1 · 2018-06-04T18:33:25

#I=3x+8ln|x-4|-2ln|x+2|+c#

Here,

#I=int(3x^2)/(x^2-2x-8)dx#

#I=int(color(red)(3x^2))/((x-4)(x+2))dx...to(A)#

Now,

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

#color(red)(2x(x+2)+x(x-4)=3x^2#

Subst. for , #color(red)(3x^2)# , into #(A)#

#I=int([color(red)(2x(x+2)+x(x-4))])/((x-4)(x+2))dx#

#=int[(2x(x+2))/((x-4)(x+2))+(x(x-4))/((x-4)(x+2))]dx#

#=2int(x)/(x-4)dx+intx/(x+2)dx#

#=2int(x-4+4)/(x-4)dx+int(x+2-2)/(x+2)dx#

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

#=2[x+4ln|x-4|]+[x-2ln|x+2|]+c#

#=2x+8ln|x-4|+x-2ln|x+2|+c#

#=3x+8ln|x-4|-2ln|x+2|+c#