Solution to Stewart Calculus
Answer
You should use the substitution rule. Let $u=e^z +z$. Then we have $$ du=(e^z +z)' dz=(e^z +1)dz$$ At $z=0$ we have $u=e^0 +0=1$ and at $z=1$ we have $u=e^1 +1=e+1$. So, after the substitution, the integral becomes $$\int _0 ^1 \frac {e^z +1}{e^z +z}\,dz=\int _1 ^{e+1}\frac{du}{u}=\ln|u|\,\Big]_1 ^{e+1} =\ln(e+1)-\ln(1)=\ln(e+1)-0=\ln(e+1)$$
Martin
828
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