old-question-I-couldn-t-find-it-x-x-dx-

Question Number 115594 by MJS_new last updated on 26/Sep/20

old question, I {couldn}^{'} t find it :

\int \sqrt{x - \sqrt{x}} d x = ?

Answered by MJS_new last updated on 26/Sep/20

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

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

= 2 \int t^{3 / 2} \sqrt{t - 1} d t =

[u = \sqrt{t - 1} \to d t = 2 \sqrt{t - 1} d u]

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

[v = u + \sqrt{u^{2} + 1} \to d u = \frac{\sqrt{u^{2} + 1}}{u + \sqrt{u^{2} + 1}} d v]

= \frac{1}{16} \int \frac{v^{12} + 2 v^{10} - v^{8} - 4 v^{6} - v^{4} + 2 v^{2} + 1}{v^{7}} d v =

= \frac{v^{6}}{96} + \frac{v^{4}}{32} - \frac{v^{2}}{32} + \frac{1}{32 v^{2}} - \frac{1}{32 v^{4}} - \frac{1}{96 v^{6}} - \frac{1}{4} \ln v =

[v = u + \sqrt{u^{2} + 1} \land u = \sqrt{t - 1} \land t = \sqrt{x}]

= \frac{1}{12} (8 x - 2 \sqrt{x} - 3) \sqrt{x - \sqrt{x}} - \frac{1}{4} \ln (\sqrt{\sqrt{x} - 1} + \sqrt[4]{x}) + C

Commented by Eric002 last updated on 27/Sep/20

n i c e w o r k i t h i n k i t s q 115498

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