let f(x)=((1+(√(1+x^2 )))/x) 1) calculate ∫_1 ^3 f(x)dx 2) determine f^(−1) (x) 3) find ∫ f^(−1) (x)dx .


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Question Number 45043 by maxmathsup by imad last updated on 07/Oct/18

$l e t f (x) = \frac{1 + \sqrt{1 + x^{2}}}{x}$ $1) c a l c u l a t e \int_{1}^{3} f (x) d x$ $2) d e t e r m i n e f^{- 1} (x)$ $3) f i n d \int f^{- 1} (x) d x .$
	Answered by MJS last updated on 08/Oct/18

	$\int \frac{1 + \sqrt{x^{2} + 1}}{x} d x =$ $[t = \sqrt{x^{2} + 1} \to d x = \frac{\sqrt{x^{2} + 1}}{x} d t]$ $= \int \frac{t}{t - 1} d t =$ $[u = t - 1 \to d t = d u]$ $= \int \frac{u + 1}{u} d u = u + \ln u = t - 1 + \ln (t - 1) =$ $= \sqrt{x^{2} + 1} - 1 + \ln ∣ \sqrt{x^{2} + 1} - 1 ∣=$ $= \ln ∣ 1 - \sqrt{x^{2} + 1} ∣ + \sqrt{x^{2} + 1} + C$ $(1) \underset{1}{\int^{3}} f (x) d x = \ln (\sqrt{10} + 2 \sqrt{5} - \sqrt{2} - 1) + (\sqrt{5} - 1) \sqrt{2} \approx 3.40060$
	Commented by maxmathsup by imad last updated on 08/Oct/18

	$t h a n k y o u s i r M J S .$
	Answered by MJS last updated on 08/Oct/18
	$(2) f(x) is defined for x≠0 range is R\[−1; 1] solving x=((1+(√(y^2 +1)))/y) gives y=((2x)/(x^2 −1)) but this is defined for x∈R\{−1; 1} especially for x∈]−1; 1[ ⇒ f^(−1) (x)=((2x)/(x^2 −1)) with x∈R\[−1; 1]$
	$(2)$ $f (x) is defined for x \neq 0$ $range is R ∖ [- 1; 1]$ $solving x = \frac{1 + \sqrt{y^{2} + 1}}{y} gives y = \frac{2 x}{x^{2} - 1} but this$ $is defined for x \in R ∖ {- 1; 1} especially for x \in] - 1; 1 [$ $\Rightarrow f^{- 1} (x) = \frac{2 x}{x^{2} - 1} with x \in R ∖ [- 1; 1]$
	Answered by MJS last updated on 08/Oct/18

	$(3)$ $\int \frac{2 x}{x^{2} - 1} d x =$ $[t = x^{2} - 1 \to d x = \frac{d t}{2 x}]$ $= \int \frac{d t}{t} = \ln t = \ln ∣ x^{2} - 1 ∣ + C$
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