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Answer 1 · 2016-09-11T17:45:25

#(10e)^x/(ln(10)+1)+e^(x+1)+C#

Expanding the integral, it becomes:

#=inte^x10^xdx+inte^(x+1)dx#

First working with the second integral, let #u=x+1# so #du=dx#:

#=inte^x10^x+inte^udu#

#=inte^x10^x+e^u#

#=inte^x10^x+e^(x+1)#

For the remaining integral, rewrite as follows:

#=int(10e)^xdx+e^(x+1)#

Notice that #10e# is just a constant. Use the rule: #inta^xdx=a^x/ln(a)+C#

#=(10e)^x/ln(10e)+e^(x+1)#

Note that #ln(10e)=ln(10)+ln(e)=ln(10)+1#:

#=(10e)^x/(ln(10)+1)+e^(x+1)+C#