How do you integrate #x^3 * sqrt(x^2 + 4) dx#?

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Answer 1 · 2018-06-12T07:51:46

The answer is #=1/5(x^2+4)^(5/2)-4/3(x^2+4)^(3/2)+C#

Perform this integral by substitution

Let #u=x^2+4#, #=>#, #du=2xdx#

Therefore,

The integral is

#I=intx^3sqrt(x^2+x)dx=intx^2sqrt(x^2+4)xdx#

#=1/2int(u-4)sqrtudu#

#=1/2int(u^(3/2)-4u^(1/2))du#

#=1/2*2/5u^(5/2)-2*2/3u^(3/2)#

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