How do you go about this question? #inte^(x+e^(x))dx#

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#inte^(x+e^(x))dx#

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Answer 1 · 2018-03-29T13:28:56

#inte^(x+e^x)"d"x=e^(e^x)+"c"#

#inte^(x+e^x)dx=inte^xe^(e^x)dx#

Let #u=e^(e^x)#. By the chain rule,

#(du)/dx=d/dxe^(e^x)=d/(de^x)e^(e^x)d/dxe^x=e^xe^(e^x)#

Substituting this into the integral gives

#inte^xe^(e^x)dx=intdu=u+"c"=e^(e^x)+"c"#