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Question Number 69637 by MJS last updated on 26/Sep/19

...now try this one: ∫(dx/(x^(1/2) −x^(1/3) −x^(1/6) ))=

$$...\mathrm{now}\:\mathrm{try}\:\mathrm{this}\:\mathrm{one}: \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{2}} −{x}^{\mathrm{1}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{6}} }= \\ $$

Answered by Kunal12588 last updated on 26/Sep/19

t=x^(1/6) ⇒dt=(1/6)x^(−5/6) dx⇒dx=6x^(5/6) dt=6t^5 dt ∫((6t^5 )/(t^3 −t^2 −t))dt=6∫(t^4 /(t^2 −t−1))dt t^4 =(t^2 −t−1)(t^2 +t+2)+3t+2 I=6∫(t^2 +t+2)dt+6∫((3t+2)/(t^2 −t−1))dt I_1 =∫((3t−2)/(t^2 −t+1))dt 3t−2=A(d/dt)(t^2 −t+1)+B ⇒A=(3/2),B=(7/2) I_1 =(3/2)∫((d(t^2 −t+1))/(t^2 −t+1))+(7/2)∫(dt/(t^2 −t+1)) I_1 =(3/2)log∣t^2 −t+1∣+(7/2)∫(dt/((t−(1/2))^2 +(3/4))) I_1 =(3/2)log∣t^2 −t+1∣+(7/2)×(1/(2(√3)))tan^(−1) (((t−(1/2))/((√3)/2)))+c I_1 =(3/2)log∣t^2 −t+1∣+(7/(4(√3)))tan^(−1) (((2t−1)/(√3)))+c I=2t^3 +3t^2 +12t+9log(t^2 −t+1)+((7(√3))/2)tan^(−1) (((2t−1)/(√3)))+c I=2x^(1/2) +3x^(1/3) +12x^(1/6) +9log(x^(1/3) −x^(1/6) +1)+7(√3)tan^(−1) (((2(√(x^(1/3) −x^(1/6) +1))−(√3))/(2x^(1/6) −1)))+c

$${t}={x}^{\mathrm{1}/\mathrm{6}} \\ $$$$\Rightarrow{dt}=\frac{\mathrm{1}}{\mathrm{6}}{x}^{−\mathrm{5}/\mathrm{6}} {dx}\Rightarrow{dx}=\mathrm{6}{x}^{\mathrm{5}/\mathrm{6}} {dt}=\mathrm{6}{t}^{\mathrm{5}} {dt} \\ $$$$\int\frac{\mathrm{6}{t}^{\mathrm{5}} }{{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}}{dt}=\mathrm{6}\int\frac{{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt} \\ $$$${t}^{\mathrm{4}} =\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right)+\mathrm{3}{t}+\mathrm{2} \\ $$$${I}=\mathrm{6}\int\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right){dt}+\mathrm{6}\int\frac{\mathrm{3}{t}+\mathrm{2}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt} \\ $$$${I}_{\mathrm{1}} =\int\frac{\mathrm{3}{t}−\mathrm{2}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\mathrm{3}{t}−\mathrm{2}={A}\frac{{d}}{{dt}}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)+{B} \\ $$$$\Rightarrow{A}=\frac{\mathrm{3}}{\mathrm{2}},{B}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{d}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}+\frac{\mathrm{7}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}{log}\mid{t}^{\mathrm{2}} −{t}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{2}}\int\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}{log}\mid{t}^{\mathrm{2}} −{t}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)+{c} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}{log}\mid{t}^{\mathrm{2}} −{t}+\mathrm{1}\mid+\frac{\mathrm{7}}{\mathrm{4}\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+{c} \\ $$$${I}=\mathrm{2}{t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{9}{log}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)+\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+{c} \\ $$$${I}=\mathrm{2}{x}^{\mathrm{1}/\mathrm{2}} +\mathrm{3}{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{12}{x}^{\mathrm{1}/\mathrm{6}} +\mathrm{9}{log}\left({x}^{\mathrm{1}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{6}} +\mathrm{1}\right)+\mathrm{7}\sqrt{\mathrm{3}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}\sqrt{{x}^{\mathrm{1}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{6}} +\mathrm{1}}−\sqrt{\mathrm{3}}}{\mathrm{2}{x}^{\mathrm{1}/\mathrm{6}} −\mathrm{1}}\right)+{c} \\ $$

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