Question-172636

Question Number 172636 by Mikenice last updated on 29/Jun/22

Answered by floor(10²Eta[1]) last updated on 29/Jun/22

I = \int \frac{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1} \sqrt{x^{2} + 1}} dx = \int \frac{dx}{\sqrt{x^{2} - 1}} - \int \frac{dx}{\sqrt{x^{2} + 1}} = I_{1} - I_{2}

I_{1} : \cosh^{2} x - \sinh^{2} x = 1

where, \sinh (x) = \frac{e^{x} - e^{- x}}{2} and \cosh (x) = \frac{e^{x} + e^{- x}}{2}

let x = \cosh (u) \Rightarrow dx = \sinh (u) du

\Rightarrow I_{1} = \int \frac{\sinh (u) du}{\sqrt{\sinh^{2} (u)}} = \int du = u + C_{1} = arcosh (x) + C_{1}

I_{2} : {tg}^{2} x + 1 = \sec^{2} x

let x = tg (u) \Rightarrow dx = \sec^{2} (u) du

I_{2} = \int \frac{\sec^{2} (u) du}{\sqrt{\sec^{2} (u)}} = \int \sec (u) du = \ln ∣ tg (u) + \sec (u) ∣ + C_{2}

= \ln ∣ x + \sqrt{x^{2} + 1} ∣ + C_{2}

\Rightarrow I = arcosh (x) - \ln ∣ x + \sqrt{x^{2} + 1} ∣ + C

Commented by BaliramKumar last updated on 30/Jun/22

I = \cosh^{- 1} (x) - \sinh^{- 1} (x) + C

Answered by Mathspace last updated on 29/Jun/22

Υ = \int \frac{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}}{\sqrt{x^{2} + 1} \sqrt{x^{2} - 1}} d x

= \int \frac{d x}{\sqrt{x^{2} - 1}} - \int \frac{d x}{\sqrt{x^{2} + 1}}

= a r c h (x) - a r g s h (x) + c

= l n (x + \sqrt{x^{2} - 1}) - l n (x + \sqrt{x^{2} - 1}) + c

= l n (\frac{x + \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}}) + c

Commented by Mathspace last updated on 30/Jun/22

Υ = l n (\frac{x + \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} + 1}}) + C