calculate-2-1-dt-t-1-t-2-

Question Number 40128 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{- 2}^{- 1} \frac{d t}{t \sqrt{1 + t^{2}}} .

Commented by math khazana by abdo last updated on 21/Jul/18

I =_{t = - u} - \int_{1}^{2} \frac{- d u}{- u \sqrt{1 + u^{2}}} = - \int_{1}^{2} \frac{d u}{u \sqrt{1 + u^{2}}}

c h a n g e m e n t u = s h (x) g i v e

\int_{1}^{2} \frac{d u}{u \sqrt{1 + u^{2}}} = \int_{a r g s h (1)}^{a r g s h (2)} \frac{c h (x) d x}{s h (x) c h x}

= \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} \frac{d x}{s h (x)} = 2 \int_{l n (1 + \sqrt{2})}^{l n (2 + \sqrt{5})} \frac{d x}{e^{x} - e^{- x}}

=_{e^{x} = t} 2 \int_{1 + \sqrt{2}}^{2 + \sqrt{5}} \frac{1}{t - \frac{1}{t}} \frac{d t}{t}

= 2 \int_{1 + \sqrt{2}}^{2 + \sqrt{5}} \frac{d t}{t^{2} - 1} = \int_{1 + \sqrt{2}}^{2 + \sqrt{5}} {\frac{1}{t - 1} - \frac{1}{t + 1}} d t

= {[l n ∣ \frac{t - 1}{t + 1} ∣]}_{1 + \sqrt{2}}^{2 + \sqrt{5}}

= l n (\frac{1 + \sqrt{5}}{3 + \sqrt{5}}) - l n (\frac{\sqrt{2}}{2 + \sqrt{2}}) \Rightarrow

I = l n (\frac{\sqrt{2}}{2 + \sqrt{2}}) - l n (\frac{1 + \sqrt{5}}{3 + \sqrt{5}}) .