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Question Number 81684 by naka3546 last updated on 14/Feb/20

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

Commented by Tony Lin last updated on 14/Feb/20

$$\int \frac{dx}{x^3+1}$$$$=\int \frac{dx}{(x+1)(x^2-x+1)}$$$$=\frac{1}{3}\int \frac{1}{x+1}dx-\frac{1}{3}\int \frac{x-2}{x^2-x+1}dx$$$$=\frac{1}{3}ln\mid x+1\mid -\frac{1}{3}[\int \frac{\frac{1}{2}(2x-1)}{x^2-x+1}dx-\frac{3}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}]$$$$=\frac{1}{3}ln\mid x+1\mid -\frac{1}{6}ln\mid x^2-x+1\mid +\sqrt{3}{tan}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$$

Answered by TANMAY PANACEA last updated on 14/Feb/20

$$\int \frac{dx}{(x+1)(x^2-x+1)}=\int \frac{a}{x+1}dx+\int \frac{bx+c}{x^2-x+1}dx$$$$now$$$$\frac{1}{x^3+1}=\frac{{ax}^2-ax+a+{bx}^2+bx+cx+c}{x^3+1}$$$$1=x^2(a+b)+x(-a+b+c)+a+c$$$$a+b=0$$$$-a+b+c=0$$$$a+c=1$$$$-a+(-a)+(1-a)=0$$$$a=\frac{1}{3}b=\frac{-1}{3}c=\frac{2}{3}$$$$\int \frac{\frac{1}{3}}{x+1}dx+\int \frac{\frac{-1}{3}x+\frac{2}{3}}{x^2-x+1}dx$$$$\frac{1}{3}\int \frac{dx}{x+1}-\frac{1}{3}\int \frac{x-2}{x^2-x+1}dx$$$$\frac{1}{3}\int \frac{dx}{x+1}-\frac{1}{6}\int \frac{2x-1-3}{x^2-x+1}dx$$$$\frac{1}{3}\int \frac{dx}{x+1}-\frac{1}{6}\int \frac{d(x^2-x+1)}{x^2-x+1}+\frac{1}{2}\int \frac{dx}{x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}$$$$\frac{1}{3}\int \frac{dx}{x+1}-\frac{1}{6}\int \frac{d(x^2-x+1)}{x^2-x+1}+\frac{1}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\frac{1}{3}ln(x+1)-\frac{1}{6}ln(x^2-x+1)+\frac{1}{2}\times \frac{2}{\sqrt{3}}{tan}^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)$$$$plscheck$$$$$$

Commented by peter frank last updated on 14/Feb/20

$$welcomagainmrtanymay,....$$$$longtime$$

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