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Question Number 94184 by MJS last updated on 17/May/20

$$\int \frac{x\sqrt[3]{x-a}}{\sqrt[3]{x-b}}dx=?$$$$\int \frac{\sqrt[3]{x-a}}{x\sqrt[3]{x-b}}dx=?$$

Commented by MJS last updated on 17/May/20

I can solve both but I want to know if there's an easier path. will post my solutions later.

Answered by  M±th+et+s last updated on 04/Jun/20

$$\int x\sqrt[3]{1-\frac{a-b}{x-b}}dx,\frac{a-b}{x-b}=u^3\Rightarrow dx=\frac{(a-b)\left(3u^2\right)}{{(1-u^3)}^2}du$$$$I=\int [\frac{a-b}{1-u^3}+b]\left[\frac{(a-b)\left(3u^2\right)}{{(1-u^3)}^2}\right]\left(u\right)du$$$$={(a-b)}^2\int \frac{3u^2}{{(1-u^3)}^3}du+b(a-b)\int \frac{3u^2}{{(1-u^3)}^2}du$$$$I_1=\frac{1}{2}\int [\left(u\right)\frac{\left(2\right)\left(3u^2\right)}{{(1-u^3)}^3}+\frac{1}{{(1-u^3)}^2}-\frac{1}{{(1-u^3)}^2}]du$$$$I_1=\frac{1}{2}\int d\left(\frac{u}{{(1-u^3)}^2}\right)-\frac{1}{2}\int \frac{du}{{(1-u^3)}^2}du$$$$I_3=\int \frac{1-u^3+u^3}{{(1-u^3)}^2}du=\int \frac{1}{1-u^3}du+\frac{1}{2}\int \frac{-2u-\frac{1}{u^2}+\frac{1}{u^2}}{{[\frac{1}{u}-u^2]}^2}du$$$$I_3=\int \frac{1}{(1-u)(1+u+u^2)}du-\frac{1}{2}\int \frac{-\frac{1}{u^2}-2u}{{[\frac{1}{u}-u^2]}^2}du-\frac{1}{2}\int \frac{1}{u^2{[\frac{1}{u}-u^2]}^2}du$$$$I_3=\int \frac{\frac{1}{3}}{1-u}du+\int \frac{\frac{1}{3}u+\frac{2}{3}}{u^2+u+1}du+\frac{1}{2(\frac{1}{u}-u^2)}-\frac{1}{2}\int \frac{1}{{(1-u^3)}^2}du$$$$I_3=\frac{1}{3}\int d(ln\mid 1-u\mid )+\frac{1}{6}\int \frac{2u+1-1}{u^2+u+1}du+\frac{2}{3}\int \frac{1}{u^2+u+1}du+\frac{1}{2(\frac{1}{u}-u^2)}-\frac{1}{2}{I}_3$$$$\frac{3}{2}{I}_3=\frac{1}{3}ln\mid 1-u\mid +\frac{1}{6}ln\mid u^2+u+1\mid +\frac{2}{3}{tan}^{-1}\left(\frac{2u+\frac{1}{2}}{\sqrt{3}}\right)+\frac{1}{2(\frac{1}{u}-u^2)}$$$$I_1=\frac{u}{{(1-u^3)}^2}-\frac{1}{2}{I}_3$$$$$$$$$$$$I_2=\frac{3}{2}\int \frac{1-u^3+3u^3-1}{{(1-u^3)}^2}du=\frac{3}{2}\int \frac{1-u^3-\left(u\right)(-3u^2)}{{(1-u^3)}^2}du-\frac{3}{2}\int \frac{1}{{(1-u^3)}^2}du$$$$I_2=\frac{3}{2}\int d\left(\frac{u}{1-u^3}\right)-\frac{3}{2}{I}_3$$$$$$$$I=I_1+I_2+c$$$$$$$$$$

Answered by mathmax by abdo last updated on 04/Jun/20

$$\mathrm{let}\mathrm{I}=\int \frac{1}{\mathrm{x}}\left({}^3\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}}\right)\mathrm{dx}\mathrm{changement}{}^3\sqrt{\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}}=\mathrm{t}\mathrm{give}\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}-\mathrm{b}}={\mathrm{t}}^3\Rightarrow$$$$\mathrm{x}-\mathrm{a}={\mathrm{t}}^3\mathrm{x}-{\mathrm{bt}}^3\Rightarrow (1-{\mathrm{t}}^3)\mathrm{x}=\mathrm{a}-{\mathrm{bt}}^3\Rightarrow \mathrm{x}=\frac{\mathrm{a}-{\mathrm{bt}}^3}{1-{\mathrm{t}}^3}=\frac{{\mathrm{bt}}^3-\mathrm{a}}{{\mathrm{t}}^3-1}$$$$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{3\mathrm{bt}}^2({\mathrm{t}}^3-1)-{3\mathrm{t}}^2({\mathrm{bt}}^3-\mathrm{a})}{{({\mathrm{t}}^3-1)}^2}=\frac{{3\mathrm{bt}}^5-{3\mathrm{bt}}^2-{3\mathrm{bt}}^5+{3\mathrm{at}}^2}{{({\mathrm{t}}^3-1)}^2}\Rightarrow$$$$\mathrm{I}=\int \frac{{\mathrm{t}}^3-1}{{\mathrm{bt}}^3-\mathrm{a}}\times \mathrm{t}\times \frac{(3\mathrm{a}-3\mathrm{b}){\mathrm{t}}^2}{{({\mathrm{t}}^3-1)}^2}\mathrm{dt}=(3\mathrm{a}-3\mathrm{b})\int \frac{{\mathrm{t}}^3}{({\mathrm{bt}}^3-\mathrm{a})({\mathrm{t}}^3-1)}\mathrm{dt}$$$$=\frac{3\mathrm{a}-3\mathrm{b}}{\mathrm{b}}\int \frac{{\mathrm{t}}^3}{({\mathrm{t}}^3-{\left({}^3\sqrt{\frac{\mathrm{a}}{\mathrm{b}}}\right)}^3)({\mathrm{t}}^3-1)}\mathrm{dt}\mathrm{let}\alpha {=}^3\sqrt{\frac{\mathrm{a}}{\mathrm{b}}}\Rightarrow$$$$\mathrm{I}=\frac{3\mathrm{a}-3\mathrm{b}}{\mathrm{b}}\int \frac{{\mathrm{t}}^3}{({\mathrm{t}}^3-{\alpha }^3)({\mathrm{t}}^3-1)}\mathrm{dt}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^3}{({\mathrm{t}}^3-{\alpha }^3)({\mathrm{t}}^3-1)}\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^3}{(\mathrm{t}-\alpha )({\mathrm{t}}^2+\alpha \mathrm{t}+{\alpha }^2)(\mathrm{t}-1)({\mathrm{t}}^2+\mathrm{t}+1)}$$$$=\frac{\mathrm{a}}{\mathrm{t}-\alpha }+\frac{\mathrm{b}}{\mathrm{t}-1}+\frac{\mathrm{ct}+\mathrm{d}}{{\mathrm{t}}^2+\alpha \mathrm{t}+{\alpha }^2}+\frac{\mathrm{et}+\mathrm{f}}{{\mathrm{t}}^2+\mathrm{t}+1}\Rightarrow$$$$\int \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{aln}\mid \mathrm{t}-\alpha \mid +\mathrm{bln}\mid \mathrm{t}-1\mid +\int \frac{\mathrm{ct}+\mathrm{d}}{{\mathrm{t}}^2+\alpha \mathrm{t}+{\alpha }^2}\mathrm{dt}+\int \frac{\mathrm{et}+\mathrm{f}}{{\mathrm{t}}^2+\mathrm{t}+1}\mathrm{dt}$$$$\mathrm{rest}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{coefficients}...\mathrm{be}\mathrm{continued}...$$

Answered by MJS last updated on 05/Jun/20

$$\int \frac{x\sqrt[3]{x-a}}{\sqrt[3]{x-b}}dx=$$$$[t=\frac{\sqrt[3]{x-a}}{\sqrt[3]{x-b}}\to dx=\frac{3}{a-b}{(x-a)}^{2/3}{(x-b)}^{4/3}]$$$$=3(a-b)\int \frac{t^3({bt}^3-a)}{{(t^3-1)}^3}dt=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=\frac{a-b}{6{(t^3-1)}^2}((a-7b)t^3+2(a+2b))t+\frac{(a-b)(a+2b)}{3}\int \frac{dt}{t^3-1}=$$$$\mathrm{and}\mathrm{I}\mathrm{have}\mathrm{shown}\mathrm{that}$$$$\int \frac{dt}{t^3-1}=\frac{1}{6}\ln \frac{{(t-1)}^2}{t^2+t+1}-\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}(2t+1)}{3}$$

Answered by MJS last updated on 05/Jun/20

$$\int \frac{\sqrt[3]{x-a}}{x\sqrt[3]{x-b}}dx=$$$$[t=\frac{\sqrt[3]{x-a}}{\sqrt[3]{x-b}}\to dx=\frac{3}{a-b}{(x-a)}^{2/3}{(x-b)}^{4/3}dt]$$$$=3(a-b)\int \frac{t^3}{(t^3-1)({bt}^3-a)}dt=$$$$[\mathrm{let}a={bc}^3\iff c=\sqrt[3]{\frac{a}{b}}]$$$$=3(c^3-1)\int \frac{t^3}{(t^3-1)(t^3-c^3)}dt=$$$$=c\int \frac{dt}{t-c}-\int \frac{dt}{t-1}-c\int \frac{t+2c}{t^2+ct+c^2}dt+\int \frac{t+2}{t^2+t+1}dt=$$$$...$$$$=\frac{1}{2}\ln \frac{t^2+t+1}{{(t-1)}^2}+\frac{c}{2}\ln \frac{{(t-c)}^2}{t^2+ct+c^2}+$$$$+\sqrt{3}\arctan \frac{\sqrt{3}(2t+1)}{3}-c\sqrt{3}\arctan \frac{\sqrt{3}(2t+c)}{3}$$

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