How do you find the indefinite integral of #int(4x(3+2sqrt(x^2-3))^5) / (sqrt(x^2-3) dx#?

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Answer 1 · 2015-10-13T00:18:24

#I = (3+2sqrt(x^2-3))^6/3+C#

#I = int (4x(3+2sqrt(x^2-3))^5)/(sqrt(x^2-3)) dx#

#x^2-3=t^2 => 2xdx=2tdt#

#I = int ((3+2sqrt(t^2))^5)/(sqrt(t^2)) 4tdt = int ((3+2t)^5)/t 4tdt#

#I = int ((3+2t)^5)/t 4tdt = 4 int (3+2t)^5 dt#

#u=3+2t => du=2dt => dt=1/2du#

#I = 2 int u^5du = u^6/3+C = (3+2t)^6/3+C#

#I = (3+2sqrt(x^2-3))^6/3+C#