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Question Number 116418 by bemath last updated on 03/Oct/20

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^5\sqrt{4+{\mathrm{x}}^2}}=?$$

Answered by MJS_new last updated on 03/Oct/20

$$\int \frac{dx}{x^5\sqrt{4+x^2}}=$$$$[x=t-\frac{1}{t}\iff t=\frac{x+\sqrt{4+x^2}}{2}\to dx=\frac{2\sqrt{4+x^2}}{x+\sqrt{4+x^2}}dt]$$$$=\int \frac{t^4}{{(t^2-1)}^5}dt=$$$$\left[{\mathrm{Ostrogeadski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{t(t^2+1)(3t^4-14t^2+3)}{128{(t^2-1)}^4}+\frac{3}{128}\int \frac{dt}{t^2-1}=$$$$=\frac{t(t^2+1)(3t^4-14t^2+3)}{128{(t^2-1)}^4}+\frac{3}{256}\ln \frac{t-1}{t+1}=$$$$=\frac{(3x^2-8)\sqrt{4+x^2}}{128x^4}+\frac{3}{256}\ln \frac{-2+\sqrt{4+x^2}}{x}+C$$

Answered by MJS_new last updated on 04/Oct/20

$$\mathrm{easier}:$$$$\int \frac{dx}{x^5\sqrt{4+x^2}}=$$$$[t=\sqrt{4+x^2}\to dx=\frac{\sqrt{4+x^2}}{x}]$$$$=\int \frac{dt}{{(t^2-4)}^3}=$$$$\left[\mathrm{Ostrogradski}\mathrm{again}\right]$$$$=\frac{t(3t^2-20)}{128{(t^2-4)}^2}+\frac{3}{128}\int \frac{dt}{t^2-4}=$$$$=\frac{t(3t^2-20)}{128{(t^2-4)}^2}+\frac{3}{512}\ln \frac{t-2}{t+2}=$$$$=\frac{(3x^2-8)\sqrt{4+x^2}}{128x^4}+\frac{3}{256}\ln \frac{-2+\sqrt{4+x^2}}{x}+C$$

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