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

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Answer 1 · 2018-03-22T08:03:32

The answer is #=1/2*3^(x^2+1)/ln3 +C#

Let #u=x^2+1#, #=>#, #du=2xdx#

Therefore,

#I=intx(3^(x^2+1))dx=1/2int3^udu#

Let #v=3^u#

#lnv=ln(3^u) = u ln3#

#v=e ^(u ln3)#

Therefore,

#I=1/2inte ^(u ln3)du #

#=1/2*e ^(u ln3)/(ln3)#

#=1/2*3^(x^2+1)/ln3 +C#