How do you find #\int \frac { 3e ^ { x } } { 3e ^ { x } + 100} d x#?

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Answer 1 · 2017-03-18T06:35:54

The answer is #=ln(3e^x+100) +C#

We perform this integral by substitution

Let #u=3e^x+100#

#du=3e^xdx#

Therefore,

#int(3e^xdx)/(3e^x+100)#

#=int(du)/u#

#=lnu#

#=ln(3e^x+100) +C#