What is the integral of #e^(2x)#?

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Answer 1 · 2018-05-02T21:06:50

The answer

#inte^(2x)*dx=1/2[e^(2x)]+c#

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#inte^(2x)*dx=1/2int2*e^(2x)*dx=1/2[e^(2x)]+c#

Answer 2 · 2018-05-03T00:29:57

#1/2e^(2x)+C#

Given: #inte^(2x) \ dx#.

We can manipulate as follows:

#inte^(2x) \ dx#

#=int1/2*2e^(2x) \ dx#

#=1/2int2e^(2x) \ dx#

Now, let #u=2x,:.du=2 \ dx,dx=(du)/2#

#=1/2int2e^u*(du)/2#

#=1/2inte^u \ du#

#=1/2e^u+C#

Replace back #u=2x# to get the final integral:

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