How do you integrate #int e^xe^x# using substitution?

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Answer 1 · 2016-11-20T14:40:13

#inte^xe^xdx#

Method 1 - Immediate Substitution

Make the substitution #u=e^x#, which implies that #du=e^xdx#, so

#inte^xe^xdx=int(e^x)(e^xdx)=intudu=u^2/2=(e^x)^2/2=e^(2x)/2+C#

Method 2 - Simplification, then Substitution

Use the rule #a^b(a^c)=a^(b+c)# to rewrite the integral as

#inte^xe^xdx=inte^(2x)dx#

Now substitute #u=2x# so #du=2dx#:

#inte^(2x)dx=1/2int(e^(2x))(2dx)=1/2inte^udu#

Since #inte^udu=e^u#:

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