How do you integrate #int x^3/sqrt(x^2+9)# using trig substitutions?

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Answer 1 · 2016-12-17T10:46:59

#=1/3sqrt(x^2+9)( x^2-18)+C#

Let #t = x^2+9#.

Then, #dt =2x dx#.

Now, the given integral becomes

#1/2int (t-9)/sqrt t dt#

#=1/2int(t^(1/2)-9t^(-1/2)) dt#

#=1/2(2/3t^(3/2)-18t^(1/2))+C#

#=1/3sqrt t(t-27)+C#

#=1/3sqrt(x^2+9)( x^2-18)+C#-