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Integration by Substitution

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Question

Evaluate : $\int \frac{1}{(x + 1)^{2}} d x$

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Solution

Put $x + 1 = t d x = d t \int \frac{1}{t^{2}} d t \int t^{- 2} d t \frac{t^{- 1}}{- 1} = - \frac{1}{t} + c = \frac{- 1}{1 + x} + c$

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