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
1.6K
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
- 938 views
- $10.00
Related Questions
- Find the coordinates of the point $(1,1,1)$ in Spherical coordinates
- Proof of P = Fv.
- Connected rates of change calculus.
- Explain partial derivatives v3
- Answer is done but need help
- In what direction the function $f(x,y)=e^{x-y}+\sin (x+y^2)$ grows fastest at point $(0,0)$?
- Minimizing the cost of building a box
- Prove the trig identity $\frac{\sin x +\tan x}{1+\sec x}=\sin x$
