How do you find the indefinite integral of #int (x^2-6x-20)/(x+5)#?

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Answer 1 · 2017-04-13T14:21:20

The answer is #=((x+5)(x-27))/2+35ln(|x+5|)+C#

We perform this integral by substitution.

Let #u=x+5#, #=>#, #dx=du#

#x=u-5#

and

#x^2-6x-20=(u-5)^2-6(u-5)-20#

#=u^2-10u+25-6u+30-20#

#=u^2-16u+35#

Therefore,

#int((x^2-6x-20)dx)/(x+5)#

#=int((u^2-16u+35)du)/u#

#=int(u-16+35/u)#

#=u^2/2-16u+35lnu#

#=(x+5)^2/2-16(x+5)+35ln(|x+5|)+C#

#=(x+5)(x+5-32)/2+35ln(|x+5|)+C#

#=((x+5)(x-27))/2+35ln(|x+5|)+C#