Integrate the following function #int_0^1(e^z+1)/(e^z+z# #dz#? Need help with u substitution?

Question

I have already asked this question (socratic link) but I'm not sure how to use u- substitution to get #[ln(e^z+z)]_0^1# . Would I start off by letting #u=e^z+z#? And then #du=(e^z+1)dx#? Which steps would I need to do next?

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Answer 1 · 2018-06-27T01:06:38

Please see below.

.

#int_0^1(e^z+1)/(e^z+z)dz#

#u=e^z+z#

#du=(e^z+1)dz#

#int_0^1(e^z+1)/(e^z+z)dz=int(du)/u=lnabsu=lnabs(e^z+z))_0^1=lnabs(e+1)-lnabs(1+0)=lnabs(e+1)-0=lnabs(e+1)#