How do you integrate #int xsqrt(x^2+9)# using substitution?

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Answer 1 · 2017-02-01T06:39:55

The answer is #=1/3(x^2+9)^(3/2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Let #u=x^2+9#

#du=2xdx#

#xdx=1/2du#

#intxsqrt(x^2+9)dx=1/2intsqrtudu#

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

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