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$$\int _{0}^ {e-1} \frac {x} {x+1}\ dx\ $$

Do I have to use substitution here?
$u = x+1,\; du = dx$

What I have already tried: $$\frac {1+1} {x+1} = \int 1 \ dx + \int \frac {1} {x+1} dx $$

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$$\int _{0}^ {e-1} \frac {x} {x+1}\ dx=\int _{0}^ {e-1} \frac {x+1-1} {x+1}\ dx \\= \int _{0}^ {e-1}1- \frac {1} {x+1}\ dx \\ =[x-\ln (x+1)] _{0}^ {e-1}$$

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You know that $\int \dfrac{1}{x+1}=\log (x+1)+C$

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