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Question Number 64227 by Tony Lin last updated on 16/Jul/19

$$\int \frac{1}{x^6+x^3}dx=?$$

Answered by ajfour last updated on 16/Jul/19

$$I=\int \frac{dx}{x^3(1+x^3)}$$$$\frac{1}{x^3(1+x^3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{1+x}+\frac{Ex+F}{x^2-x+1}$$$$\Rightarrow 1=(1+x^3)({Ax}^2+Bx+C)$$$$+{Dx}^3(x^2-x+1)+(Ex+F)x^3(1+x)$$$$x=0\Rightarrow C=1$$$$x=-1\Rightarrow D=-\frac{1}{3}$$$$\Rightarrow A+D+E=0$$$$B-D+E+F=0$$$$C+D+F=0$$$$A=0,B=0$$$$\Rightarrow E=\frac{1}{3},F=-\frac{2}{3}$$$$\Rightarrow I=\int \frac{dx}{x^3}-\frac{1}{3}\int \frac{dx}{1+x}+\frac{1}{3}\int \frac{(x-2)dx}{x^2-x+1}$$$$=-\frac{1}{2x^2}-\frac{1}{3}\ln (1+x)+\frac{1}{6}\ln (x^2-x+1)$$$$-\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$I=-\frac{1}{2x^2}-\frac{1}{3}\ln (1+x)+\frac{1}{6}\ln (x^2-x+1)$$$$-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c.$$

Commented by Tony Lin last updated on 16/Jul/19

$$thankssir$$

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