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

Question Number 46594 by maxmathsup by imad last updated on 29/Oct/18

f i n d \int (\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}}) d x

Answered by MJS last updated on 29/Oct/18

\int \sqrt{x + \sqrt{x}} d x =

[t = \sqrt{x} \to d x = 2 \sqrt{x} d t]

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

\int (2 t + 1) \sqrt{t^{2} + t} d t = \frac{2}{3} {(t^{2} + t)}^{\frac{3}{2}} = \frac{2}{3} {(x + \sqrt{x})}^{\frac{3}{2}}

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

[u = 2 t + 1 \to d t = \frac{1}{2} d u]

= - \frac{1}{4} \int \sqrt{u^{2} - 1} d u =

[v = arccosh u \to d u = \sinh v d v]

= - \frac{1}{4} \int \sinh^{2} v d v = \frac{1}{8} \int d v - \frac{1}{8} \int \cosh 2 v d v =

= \frac{1}{8} v - \frac{1}{16} \sinh 2 v = \frac{1}{8} arccosh u - \frac{1}{16} \sinh (2 arccosh u) =

= \frac{1}{8} arccosh (2 t + 1) - \frac{1}{16} \sinh (2 arccosh (2 t + 1)) =

= \frac{1}{8} arccosh (2 \sqrt{x} + 1) - \frac{1}{16} \sinh (2 arccosh (2 \sqrt{x} + 1))

\int \sqrt{x + \sqrt{x}} d x = \frac{2}{3} {(x + \sqrt{x})}^{\frac{3}{2}} + \frac{1}{8} arccosh (2 \sqrt{x} + 1) - \frac{1}{16} \sinh (2 arccosh (2 \sqrt{x} + 1)) + C

similar

- \int \sqrt{x - \sqrt{x}} d x = - \frac{2}{3} {(x - \sqrt{x})}^{\frac{3}{2}} + \frac{1}{8} arccosh (2 \sqrt{x} - 1) - \frac{1}{16} \sinh (2 arccosh (2 \sqrt{x} - 1)) + C

Commented by maxmathsup by imad last updated on 29/Oct/18

g o o d w o r k i s d o n e t h a n k y o u s i r M J S .