p-p-1-dp-

Question Number 10147 by ridwan balatif last updated on 27/Jan/17

\int \sqrt{\frac{p}{p - 1}} dp = \dots ?

Commented by prakash jain last updated on 27/Jan/17

p = x^{2}

d p = 2 x d x

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

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

substitute x = \cosh t

d x = \sinh t d t

= \int {2 \cosh}^{2} t d t

= \int (1 + \cosh 2 t) d t

= (t + \frac{1}{2} \sinh 2 t) + C

= (t + \sinh t \cosh t) + C

= (\cosh^{- 1} x + x \sqrt{x^{2} - 1}) + C

= \ln (x + \sqrt{x^{2} - 1}) + x \sqrt{x^{2} - 1} + C

= \ln (\sqrt{p} + \sqrt{p - 1}) + \sqrt{p (p - 1)} + C