find-2-3-x-1-x-1-x-dx-

Question Number 30182 by abdo imad last updated on 17/Feb/18

f i n d \int_{2}^{3} \frac{\sqrt{x + 1}}{x \sqrt{1 - x}} d x .

Commented by abdo imad last updated on 21/Feb/18

l e t p u t \sqrt{\frac{x + 1}{x - 1}} = t \Leftrightarrow \frac{x + 1}{x - 1} = t^{2} \Leftrightarrow x + 1 = - t^{2} + t^{2} x \Leftrightarrow

(1 - t^{2}) x = - t^{2} - 1 \Leftrightarrow x = \frac{- t^{2} - 1}{- t^{2} + 1} = \frac{t^{2} + 1}{t^{2} - 1} = 1 + \frac{2}{t^{2} - 1} \Rightarrow

\frac{d x}{d t} = \frac{- 4 t}{{(t^{2} - 1)}^{2}} \Rightarrow I = \int_{\sqrt{3}}^{\sqrt{2}} t \frac{t^{2} - 1}{t^{2} + 1} \frac{- 4 t}{{(t^{2} - 1)}^{2}} d t

= 4 \int_{\sqrt{2}}^{\sqrt{3}} \frac{t^{2}}{(t^{2} + 1) (t^{2} - 1)} d t = 4 \int_{\sqrt{2}}^{\sqrt{3}} \frac{t^{2} - 1 + 1}{(t^{2} + 1) (t^{2} - 1)} d t

= 4 \int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{t^{2} + 1} + 4 \int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{(t^{2} + 1) (t^{2} - 1)} w e h a v e

\int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{1 + t^{2}} = a r c t a n (\sqrt{3}) - a r c t a n (\sqrt{2}) a n d

\int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{(t^{2} + 1) (t^{2} - 1)} = \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{3}} (\frac{1}{t^{2} - 1} - \frac{1}{t^{2} + 1}) d t

= \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{t^{2} - 1} - \frac{1}{2} (a r c t a n (\sqrt{3)} - a r c t a n \sqrt{2}))

= \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{3}} (\frac{1}{t - 1} - \frac{1}{t + 1}) d t - \frac{1}{2} (\dots .)

= \frac{1}{2} {[l n ∣ \frac{t - 1}{t + 1} ∣]}_{\sqrt{2}}^{\sqrt{3}} - \frac{1}{2} (\dots) \Rightarrow

I = 4 (a r t a n (\sqrt{3}) - a r c t a n (\sqrt{2})) + 2 \int_{\sqrt{2}}^{\sqrt{3}} \frac{d t}{t^{2} - 1} d t

- 2 (a r c t a n (\sqrt{3}) - a r c t a n (\sqrt{2}))

= 2 (a r c t a n (\sqrt{3}) - a r c t a n (\sqrt{2})) + l n ∣ \frac{\sqrt{3} - 1}{\sqrt{3} + 1} ∣ - l n ∣ \frac{\sqrt{2} - 1}{\sqrt{2} + 1} ∣ .

Commented by abdo imad last updated on 21/Feb/18

f o r g i v e \dots t h e Q i s f i n d \int_{2}^{3} \frac{\sqrt{x + 1}}{x \sqrt{x - 1}} d x .