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

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

Answered by Kunal12588 last updated on 26/Sep/19

$$t=x^{1/6}$$$$\Rightarrow dt=\frac{1}{6}{x}^{-5/6}dx\Rightarrow dx=6x^{5/6}dt=6t^{5}dt$$$$\int \frac{6t^5}{t^3-t^2-t}dt=6\int \frac{t^4}{t^2-t-1}dt$$$$t^4=(t^2-t-1)(t^2+t+2)+3t+2$$$$I=6\int (t^2+t+2)dt+6\int \frac{3t+2}{t^2-t-1}dt$$$$I_1=\int \frac{3t-2}{t^2-t+1}dt$$$$3t-2=A\frac{d}{dt}(t^2-t+1)+B$$$$\Rightarrow A=\frac{3}{2},B=\frac{7}{2}$$$$I_1=\frac{3}{2}\int \frac{d(t^2-t+1)}{t^2-t+1}+\frac{7}{2}\int \frac{dt}{t^2-t+1}$$$$I_1=\frac{3}{2}log\mid t^2-t+1\mid +\frac{7}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2+\frac{3}{4}}$$$$I_1=\frac{3}{2}log\mid t^2-t+1\mid +\frac{7}{2}\times \frac{1}{2\sqrt{3}}{tan}^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$$$$I_1=\frac{3}{2}log\mid t^2-t+1\mid +\frac{7}{4\sqrt{3}}{tan}^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+c$$$$I=2t^3+3t^2+12t+9log(t^2-t+1)+\frac{7\sqrt{3}}{2}{tan}^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+c$$$$I=2x^{1/2}+3x^{1/3}+12x^{1/6}+9log(x^{1/3}-x^{1/6}+1)+7\sqrt{3}{tan}^{-1}\left(\frac{2\sqrt{x^{1/3}-x^{1/6}+1}-\sqrt{3}}{2x^{1/6}-1}\right)+c$$

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