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Question Number 191243 by sonukgindia last updated on 21/Apr/23

Commented by mr W last updated on 21/Apr/23

$$atx=1?$$

Answered by PandaMa last updated on 21/Apr/23

$$atx=1thereispointofsingularity.thereforethatintegraldoesnotconverge.sowecannotcalculateit$$$$$$

Answered by mr W last updated on 22/Apr/23

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

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