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Integration by Substitution

∫ tan - 1 x +...

Question

$\int t a n^{- 1} (x + 1) + C d x$

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Solution

$\begin{matrix} I_{1} = \int {tan}^{- 1} (x + 1) + C p u t (x + 1) = t d x = d t I_{1} = \int 1. {tan}^{- 1} (t) d t = t . {tan}^{- 1} - \int \frac{t}{1 + t^{2}} d t = t . {tan}^{- 1} (t) - \frac{1}{2} log (1 + t^{2}) + C = (x + 1) {tan}^{- 1} (x + 1) - \frac{1}{2} log [1 + {(1 + x)}^{2}] + C \end{matrix}$

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