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

Question Number 148812 by mathmax by abdo last updated on 31/Jul/21

find \int \frac{dx}{(\sqrt{x} + \sqrt{x + 1}) (\sqrt{x - 1} + \sqrt{x})}

Answered by MJS_new last updated on 31/Jul/21

\int \frac{d x}{(\sqrt{x} + \sqrt{x + 1}) (\sqrt{x} + \sqrt{x - 1})} =

= \int (\frac{1}{(\sqrt{x} + \sqrt{x + 1}) (\sqrt{x} + \sqrt{x - 1})} \times \frac{(\sqrt{x} - \sqrt{x + 1}) (\sqrt{x} - \sqrt{x - 1})}{(\sqrt{x} - \sqrt{x + 1}) (\sqrt{x} - \sqrt{x - 1})}) d x =

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

[which is easy to solve]

= \frac{x^{2}}{2} + \frac{x \sqrt{x^{2} - 1} - \ln (x + \sqrt{x^{2} - 1})}{2} - \frac{2 (2 x - 1) \sqrt{x (x - 1)} - \ln (2 x - 1 + 2 \sqrt{x (x - 1)})}{8} - \frac{2 (2 x + 1) \sqrt{x (x + 1)} - \ln (2 x + 1 + 2 \sqrt{x (x + 1)})}{8} + C

Commented by MJS_new last updated on 01/Aug/21

{you}^{'} re welcome as always and thanks for your

concern; I^{'} m well . Maybe a bit worn out; -)

Commented by mathmax by abdo last updated on 01/Aug/21

thank sir mjs you are absent for a time i hope that your are in a

good helth \dots .