# Use a table of integrals to evaluate the following integrals. int x/(x+1)dx

Use a table of integrals to evaluate the following integrals. $\int \frac{x}{x+1}dx$
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vyhlodatis
$\int \frac{x}{x+1}dx$
Formula: $\int \frac{udu}{a+bu}=\frac{1}{{b}^{2}}\left(a+bu-a\mathrm{ln}|a+bu|\right)+C$
$u=x⇒du=x$, and a=1, b=1
Therefore,
$\int \frac{x}{x+1}dx=\underset{\int \frac{udu}{a+bu}}{\underset{⏟}{\int \frac{x}{1+1x}dx}}=\frac{1}{\left(1{\right)}^{2}}\left(1+1x-1\mathrm{ln}|1+1x|\right)+C$
Simplify
$\int \frac{x}{x+1}dx=1+x-\mathrm{ln}|1+x|+C$
$\int \frac{x}{x+1}dx=x-\mathrm{ln}|x+1|+C$
Result:
$x-\mathrm{ln}|x+1|+C$