Solve : $\int \frac{1}{x + 1} d x$

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Solution

$I = \int \frac{1}{x + 1} d x$

Let $x + 1 = t$
$d x = d t$
Therefore,
$I = \int \frac{d t}{t}$

$I = log t + C$
$I = log (x + 1) + C$

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