In this case, the integrand is the derivative of a function composition (up to a multiplicative constant):

#g(y)=e^y#

#f(x)=x^2+1#

#rArr g(f(x))=e^(x^2+1)#

#rArr d/(dx) [g(f(x))]=d/(dx)[e^(x^2+1)]=2x e^(x^2+1)#

The only difference is the multiplicative constant (it's #2# instead of #3#), so we can rewrite the integral to look like the derivative of #e^(x^2+1)# by working with multiplicative constants:

#int 3x e^(x^2+1) dx=3/2 int 2x e^(x^2+1) dx=3/2 e^(x^2+1) + C#

Another (very similar) way of solving this was to consider the substitution #u(x)=x^2+1#. We could relate the two differentials #du=2x dx# and rewrite the integral in terms of #u#:

#int 3x e^(x^2+1) dx=int 3/2 e^(x^2+1) (2xdx)=int 3/2 e^u du=3/2e^u+C#

Now we could return back to #x#:

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