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Question Number 201110 by emilagazade last updated on 29/Nov/23

$$\int \frac{1}{\sqrt{{(x-a)}^3}+\sqrt{{(x+a)}^3}}dx$$

Answered by Frix last updated on 29/Nov/23

$$\sqrt{p}+\sqrt{q}=\sqrt{p+2\sqrt{pq}+q}$$$$\int \frac{dx}{{(x-a)}^{\frac{3}{2}}+{(x+a)}^{\frac{3}{2}}}=$$$$=\frac{1}{\sqrt{2}}\int \frac{dx}{{(x(x^2+3a^2)+{(x^2-a^2)}^{\frac{3}{2}})}^{\frac{1}{3}}}=$$$$(\mathrm{use}t=\frac{x+\sqrt{x^2-a^2}}{a}\iff x=\frac{a(t^2+1)}{2t}\to dx=\frac{a(t^2-1)}{2t^2}dt)$$$$=\frac{1}{\sqrt{2a}}\int \frac{t^2-1}{t^{\frac{3}{2}}(t^2+3)}dt\stackrel{u=\sqrt{t}}{=}\frac{\sqrt{2}}{\sqrt{a}}\int \frac{u^4-1}{u^2(u^4+3)}du=$$$$(\mathrm{decompose}...)$$$$=\frac{\sqrt{2}}{3\sqrt{a}u}+\frac{2}{3^{\frac{5}{4}}\sqrt{a}}({\tan }^{-1}(\frac{\sqrt{2}u}{3^{\frac{1}{4}}}+1)+{\tan }^{-1}(\frac{\sqrt{2}u}{3^{\frac{1}{4}}}-1))+\frac{1}{3^{\frac{5}{4}}\sqrt{a}}\ln \frac{u^2-3^{\frac{1}{4}}\sqrt{2}u+\sqrt{3}}{u^2+3^{\frac{1}{4}}\sqrt{2}u+\sqrt{3}}$$$$\mathrm{Now}\mathrm{insert}$$$$u=\frac{\sqrt{x+\sqrt{x^2-a^2}}}{\sqrt{a}}$$

Commented by emilagazade last updated on 29/Nov/23

$$thankyoualotSir$$

Commented by Frix last updated on 29/Nov/23

$$\mathrm{I}\mathrm{also}\mathrm{tried}$$$$\frac{1}{{(x-a)}^{\frac{3}{2}}+{(x+a)}^{\frac{3}{2}}}=\frac{{(x+a)}^{\frac{3}{2}}-{(x-a)}^{\frac{3}{2}}}{2a(3x^2+a^2)}$$$$\mathrm{leading}\mathrm{to}$$$$\frac{1}{2a}(\int \frac{{(x+a)}^{\frac{3}{2}}}{3x^2+a^2}dx-\int \frac{{(x-a)}^{\frac{3}{2}}}{3x^2+a^2}dx)$$$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{better}...$$

Commented by emilagazade last updated on 29/Nov/23

$$thisoneItriedtoo.butthebooksuggeststodosubstitutionwhichleadstopartialfractions.$$

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