∫^ (1/(1+(√(1+t)) )) dt=?


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Question Number 142100 by ZiYangLee last updated on 26/May/21

$\int^{} \frac{1}{1 + \sqrt{1 + t}} d t = ?$
	Answered by iloveisrael last updated on 26/May/21

	$I = \int \frac{d t}{1 + \sqrt{1 + t}} = \int \frac{1 - \sqrt{1 + t}}{- t} d t$ $I = \int \frac{\sqrt{1 + t}}{t} d t - \ln t + c$ $l e t \sqrt{1 + t} = u \Rightarrow t = u^{2} - 1$ $d t = 2 u d u$ $I = \int \frac{u}{u^{2} - 1} (2 u d u) - \ln t + c$ $I = 2 \int \frac{u^{2} - 1 + 1}{u^{2} - 1} d u - \ln t + c$ $I = 2 u + \int (\frac{1}{u - 1} - \frac{1}{u + 1}) d u - \ln t + c$ $I = 2 \sqrt{1 + t} + \ln ∣ \frac{u - 1}{u + 1} ∣ - \ln t + C$ $I = 2 \sqrt{1 + t} + \ln ∣ \frac{\sqrt{1 + t} - 1}{t (\sqrt{1 + t} + 1)} ∣ + C$
	Answered by mathmax by abdo last updated on 26/May/21

	$I = \int \frac{dt}{1 + \sqrt{1 + t}} we do the changement 1 + \sqrt{1 + t} = x \Rightarrow$ $\sqrt{1 + t} = x - 1 \Rightarrow 1 + t = {(x - 1)}^{2} \Rightarrow t = x^{2} - 2 x + 1 - 1 = x^{2} - 2 x \Rightarrow$ $I = \int \frac{2 x - 2}{x} dx = 2 x - 2 \log ∣ x ∣ + C$ $= 2 (1 + \sqrt{1 + t}) - 2 \log ∣ 1 + \sqrt{1 + t} ∣ + C$ $= 2 \sqrt{1 + t} - 2 \log (1 + \sqrt{1 + t}) + K$
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