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I want to calculate the integral $$\int\frac{x-1}{x+1}\,\mathrm{d}x.$$ I have tried solving it by differentiating the denominator and substituting it, but I didn't get it. How else can I solve it?

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    $\begingroup$ First, do the division. $\endgroup$ – David Mitra Aug 23 '15 at 12:13
  • $\begingroup$ How do I do that? $\endgroup$ – High Zedd Aug 23 '15 at 12:15
  • $\begingroup$ $$\frac{x-1}{x+1}=\frac{x}{x+1}-\frac{1}{x+1}$$ $\endgroup$ – ParaH2 Aug 23 '15 at 12:16
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    $\begingroup$ @HighZedd $$\frac{x-1}{x+1} = \frac{x+1-2}{x+1} = 1-\frac2{x+1}$$ $\endgroup$ – peterwhy Aug 23 '15 at 12:16
  • $\begingroup$ $\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}$ $\endgroup$ – Claude Leibovici Aug 23 '15 at 12:16
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$$\begin{align*} \int\frac{x-1}{x+1}\ dx &= \int\frac{x+1-2}{x+1}\ dx\\ &= \int\left(1-\frac2{x+1}\right)\ dx\\ &= \int dx - 2\int\frac{dx}{x+1}\\ &= \cdots \end{align*}$$

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