I = ∫_0 ^4 tanh^(−1) 2x dx = ???


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Question Number 125662 by physicstutes last updated on 12/Dec/20

$I = \underset{0}{\int^{4}} \tanh^{- 1} 2 x d x = ? ? ?$
	Answered by mathmax by abdo last updated on 12/Dec/20

	$th (x) = y \Rightarrow y = \frac{sh (x)}{ch (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} = \frac{e^{2 x} - 1}{e^{2 x} + 1}$ $let e^{2 x} = u \Rightarrow \frac{u - 1}{u + 1} = y \Rightarrow u - 1 = yu + y \Rightarrow (1 - y) u = 1 + y \Rightarrow u = \frac{1 + y}{1 - y}$ $\Rightarrow e^{2 x} = \frac{1 + y}{1 - y} \Rightarrow 2 x = \ln (\frac{1 + y}{1 - y}) \Rightarrow x = \frac{1}{2} \ln (\frac{1 + x}{1 - x}) = {th}^{- 1} (y) \Rightarrow$ ${th}^{- 1} (x) = \frac{1}{2} \ln (\frac{1 + x}{1 - x}) \Rightarrow I = \frac{1}{2} \int_{0}^{4} \ln (\frac{1 + 2 x}{1 - 2 x}) dx$ $we do the changement \frac{1 + 2 x}{1 - 2 x} = t \Rightarrow 1 + 2 x = t - 2 tx \Rightarrow$ $(2 + 2 t) x = t - 1 \Rightarrow x = \frac{t - 1}{2 (t + 1)} \Rightarrow \frac{dx}{dt} = \frac{1}{2} \times \frac{t + 1 - (t - 1)}{{(t + 1)}^{2}} = \frac{1}{{(t + 1)}^{2}} \Rightarrow$ $I = \frac{1}{2} \int_{1}^{- \frac{9}{7}} \ln (t) \frac{dt}{{(t + 1)}^{2}} \Rightarrow I is not defined . . . .! :$
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