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

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Answer 1 · 2018-04-25T15:53:43

The answer #(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+c#

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#intx^3*(x^2+1)^(1/2)*dx#

suppose:
#u=x^2+1#
#x=(u-1)^(1/2)#
#du=2x*dx#
#dx=1/(2x)*du#
the int become after suppose:

#int[sqrt(u-1)]^3*sqrt(u)*1/(2*sqrt(u-1))*du#

#1/2int[sqrt(u-1)]^2*sqrtu*du#

#1/2int(u-1)*sqrtu*du=1/2intu*sqrtu-sqrtu*du=1/2intu^(3/2)-u^(1/2)*du#

#1/2[2/5*u^(5/2)-2/3*u^(3/2)]+c#

#u^(5/2)/5-u^(3/2)/3+c#

#(x^2+1)^(5/2)/5-(x^2+1)^(3/2)/3+c#