find-x-1-x-1-dx-

Question Number 47062 by maxmathsup by imad last updated on 04/Nov/18

f i n d \int \sqrt{\frac{\sqrt{x} - 1}{\sqrt{x} + 1}} d x

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

t^{2} = x d x = 2 t d t

\int \sqrt{\frac{t - 1}{t + 1}} \times 2 t d t

\int \frac{2 t (t - 1)}{\sqrt{t^{2} - 1}} d t

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

2 \int \sqrt{t^{2} - 1} + 2 \int \frac{d t}{\sqrt{t^{2} - 1}} - \int \frac{d (t^{2} - 1)}{\sqrt{t^{2} - 1}}

2 [\frac{t}{2} \sqrt{t^{2} - 1} - \frac{1^{2}}{2} l n ∣ t + \sqrt{t^{2} - 1}] + 2 l n ∣ t + \sqrt{t^{2} - 1} ∣ - \frac{{(t^{2} - 1)}^{\frac{1}{2}}}{\frac{1}{2}} + c

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

Commented by maxmathsup by imad last updated on 05/Nov/18

t h a n k y o u s i r T a n m a y .