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Question Number 118959 by bramlexs22 last updated on 21/Oct/20

$$\int \frac{dx}{x^6-x^3}?$$

Answered by Dwaipayan Shikari last updated on 21/Oct/20

$$\int \frac{1}{x^3-1}-\int \frac{1}{x^3}$$$$=\int \frac{1}{x}(\frac{1}{x-1}-\frac{x-1}{x^2+x+1})+\frac{1}{2x^2}dx$$$$=\int \frac{1}{x(x-1)}-\frac{x-1}{x(x^2+x+1)}dx+\frac{1}{2x^2}$$$$=log\left(\frac{x-1}{x}\right)-\int \frac{1}{x^2+x+1}+\int \frac{1}{x(x^2+x+1)}+\frac{1}{2x^2}$$$$=log\left(\frac{x-1}{x}\right)+\frac{1}{2x^2}+\int \frac{1}{x}-\frac{x+1}{x^2+x+1}-\int \frac{1}{x^2+x+1}dx$$$$=log(x-1)+\frac{1}{2x^2}-\int \frac{x+2}{x^2+x+1}dx$$$$=log(x-1)+\frac{1}{2x^2}-\frac{1}{2}\int \frac{2x+1}{x^2+x+1}-\frac{3}{2}\int \frac{1}{x^2+x+1}$$$$=log(x-1)-\frac{1}{2}log(x^2+x+1)+\frac{1}{2x^2}-\frac{3}{2}\int \frac{1}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$$$$=log\left(\frac{x-1}{\sqrt{x^2+x+1}}\right)+\frac{1}{2x^2}-\frac{1}{\sqrt{3}}{tan}^{-1}\frac{2x+1}{\sqrt{3}}+C$$

Answered by 1549442205PVT last updated on 21/Oct/20

$$\frac{1}{{\mathrm{x}}^6-{\mathrm{x}}^3}=\frac{1}{{\mathrm{x}}^3({\mathrm{x}}^3-1)}=\frac{1}{{\mathrm{x}}^3-1}-\frac{1}{{\mathrm{x}}^3}=\frac{1}{{\mathrm{x}}^3-1}-\frac{1}{{\mathrm{x}}^3}$$$$\frac{1}{{\mathrm{x}}^3-1}=\frac{1}{(\mathrm{x}-1)({\mathrm{x}}^2+\mathrm{x}+1)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{bx}+\mathrm{c}}{{\mathrm{x}}^2+\mathrm{x}+1}$$$$\iff 1\equiv (\mathrm{a}+\mathrm{b}){\mathrm{x}}^2+(\mathrm{a}-\mathrm{b}+\mathrm{c})\mathrm{x}+\mathrm{a}-\mathrm{c}$$$$\iff \{\begin{array}{l}\mathrm{a}+\mathrm{b}=0\\ \mathrm{a}-\mathrm{b}+\mathrm{c}=0\\ \mathrm{a}-\mathrm{c}=1\end{array}\Rightarrow \{\begin{array}{l}2\mathrm{a}+\mathrm{c}=0\\ \mathrm{a}-\mathrm{c}=1\end{array}\iff \{\begin{array}{l}\mathrm{a}=1/3=-\mathrm{b}\\ \mathrm{c}=-2/3\end{array}$$$$\frac{1}{{\mathrm{x}}^3-1}=\frac{1}{3(\mathrm{x}-1)}-\frac{\mathrm{x}+2}{3({\mathrm{x}}^2+\mathrm{x}+1)}$$$$\mathrm{F}=\int \frac{dx}{x^6-x^3}=\frac{1}{3}\int \frac{\mathrm{dx}}{\mathrm{x}-1}-\int \frac{\mathrm{dx}}{{\mathrm{x}}^3}$$$$-\frac{1}{3}.(\frac{1}{2}\int \frac{2\mathrm{x}+1}{{\mathrm{x}}^2+\mathrm{x}+1}+\frac{3}{2}.\frac{1}{{\mathrm{x}}^2+\mathrm{x}+1})\mathrm{dx}$$$$=\frac{1}{3}\ln \mid \mathrm{x}-1\mid +\frac{1}{{2\mathrm{x}}^2}-\frac{1}{6}\ln ({\mathrm{x}}^2+\mathrm{x}+1)$$$$-\frac{1}{2}\int \frac{\mathrm{dx}}{{(\mathrm{x}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$=\frac{1}{3}\ln \mid \mathrm{x}-1\mid +\frac{1}{{2\mathrm{x}}^2}-\frac{1}{6}\ln ({\mathrm{x}}^2+\mathrm{x}+1)$$$$-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{C}$$$$$$

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