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 $$\int _0 ^1 \frac {e^z +1}{e^z +z}\,dz=\int _1 ^{e+1}\frac{du}{u}=\ln|u|\,\Big]_1 ^{e+1} =\ln(e+1)-\ln(1)=\ln(e+1)-0=\ln(e+1)$$
1.7K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- accepted
- 2439 views
- $10.00
Related Questions
- The cross sectional area of a rod has a radius that varies along its length according to the formula r = 2x. Find the total volume of the rod between x = 0 and x = 10 inches.
- Help with differentating business caluclus problem.
- Evaluate $\int \ln(\sqrt{x+1}+\sqrt{x}) dx$
- Let $z = f(x − y)$. Show that $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=0$
- Calculus - Derivatives (help with finding a geocache)
- Calculus Questions - Domains; Limits; Derivatives; Integrals
- Create a function whose derivate is:
- Two persons with the same number of acquaintance in a party