calculate-dx-2-x-2-3-x-2-

Question Number 138128 by mathmax by abdo last updated on 10/Apr/21

calculate \int \frac{dx}{\sqrt{2 - x^{2}} + \sqrt{3 + x^{2}}}

Answered by MJS_new last updated on 10/Apr/21

\int \frac{d x}{\sqrt{2 - x^{2}} + \sqrt{3 + x^{2}}} = \int \frac{\sqrt{x^{2} + 3}}{2 x^{2} + 1} d x - \int \frac{\sqrt{2 - x^{2}}}{2 x^{2} + 1} d x

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

[u = \frac{x + \sqrt{x^{2} + 3}}{\sqrt{3}} \to d x = \frac{\sqrt{3 (x^{2} + 3)}}{x + \sqrt{x^{2} + 3}} d u]

= \frac{3}{2} \int \frac{{(u^{2} + 1)}^{2}}{u (3 u^{4} - 4 u^{2} + 3)} d u =

= \frac{1}{2} \int \frac{d u}{u} + \frac{\sqrt{30}}{12} \int \frac{d u}{u^{2} - \frac{\sqrt{30}}{3} u + 3} - \frac{\sqrt{30}}{12} \int \frac{d u}{u^{2} + \frac{\sqrt{30}}{3} u + 3} =

= \frac{1}{2} \ln u + \frac{\sqrt{5}}{2} \arctan (\sqrt{6} u - \sqrt{5}) - \frac{\sqrt{5}}{2} \arctan (\sqrt{6} u + \sqrt{5}) =

= \frac{1}{2} \ln (x + \sqrt{x^{2} + 3}) + \frac{\sqrt{5}}{2} (\arctan (\sqrt{2} x - \sqrt{5} + \sqrt{2 (x^{2} + 3)}) - \arctan (\sqrt{2} x + \sqrt{5} + \sqrt{2 (x^{2} + 3)})) [1]

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

[v = \frac{x}{\sqrt{2 - x^{2}}} \to d x = \frac{{(2 - x^{2})}^{3 / 2}}{2} d v]

= 2 \int \frac{d v}{(v^{2} + 1) (5 v^{2} + 1)} = \frac{1}{2} \int \frac{d v}{v^{2} + \frac{1}{5}} - \frac{1}{2} \int \frac{d v}{v^{2} + 1} =

= \frac{\sqrt{5}}{2} \arctan \sqrt{5} v - \frac{1}{2} \arctan v =

= \frac{\sqrt{5}}{2} \arctan \frac{\sqrt{5} x}{\sqrt{2 - x^{2}}} - \frac{1}{2} \arctan \frac{x}{\sqrt{2 - x^{2}}} [2]

answer is [1] + [2] + C

Commented by mathmax by abdo last updated on 10/Apr/21

thank you sir mjs