How do you integrate # (x^3)((x^2 + 4)^(1/2)) dx#?

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Answer 1 · 2016-05-06T03:15:29

#=(1/15)(x^2+4)^(3/2)(3x^2+7)+C#

Use the substitution x^2+4=t^2#,

Then 2x dx =2t dt. So, #dx=(t/x)dt=t/sqrt((t^2-4)) dt#

Now, #int x^3sqrt(x^2+4) dx =int (t^2-4)^(3/2)t^2/sqrt(t^2-4)dt#

#=int(t^2-4)t^2dt=int(t^4-4t^2)dt#

#=t^5/5-t^3/3+ C#

#=t^3/15(3t^2-5)+C#

#=(1/15)(x^2+4)^(3/2)(3(x^2+4)-5)+C#

#=(1/15)(x^2+4)^(3/2)(3x^2+7)+C#