# another definite integral

I'm having trouble solving this equation..

$$\int_0^{e-1}\frac{1}{x+1}\,\mathrm dx$$

I have tried the substitution method with $u= x+1$ and $du= 1dx$

$$\int_\ udu$$

then I get $$\left.\frac{u^2}{2}\right|_0^{e-1}$$ then I change the $u$ to $x+1$

$$\left.\frac{(x+1)^2}{2}\right|_0^{e-1}$$

When I solve for $F(e-1)-F(0)$, I am stuck.

Did I set this equation correctly? Please help me understand how to solve this equation? My professor has provided me with the answer but I don't know how he had gotten the answer. The answer to the equation is $1$.

$\displaystyle \int_{0}^{e-1} \dfrac{1}{1+x}dx = \ln (1+x)|_{0}^{e-1} = \ln (1+e-1) - \ln (1+0) = \ln e - 0 = 1 - 0 = 1$
• note that the anti derivative of $\dfrac{1}{1+x}$ is $\ln(1+x)$ while the anti derivative of $x$ is $\dfrac{x^2}{2}$. – DeepSea Nov 16 '14 at 7:52