int frac{1}{x - x^3} , dx = | vedclass

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(D) To evaluate the integral $I = \int \frac{1}{x - x^3} \, dx$,we first factor the denominator: $x - x^3 = x(1 - x^2) = x(1 - x)(1 + x)$. Using parti

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$\frac{1}{2} \log \frac{x^2}{1 - x^2} + c$

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(D) To evaluate the integral $I = \int \frac{1}{x - x^3} \, dx$,we first factor the denominator: $x - x^3 = x(1 - x^2) = x(1 - x)(1 + x)$. Using partial fractions: $\frac{1}{x(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x}$. Solving for constants,we get $A = 1$,$B = 1/2$,and $C = -1/2$. Thus,$\frac{1}{x - x^3} = \frac{1}{x} + \frac{1}{2(1 - x)} - \frac{1}{2(1 + x)} = \frac{1}{x} + \frac{1}{2} \left( \frac{1}{1 - x} - \frac{1}{1 + x} \right)$. Integrating term by term: $I = \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{1 - x} \, dx - \frac{1}{2} \int \frac{1}{1 + x} \, dx$. $I = \log |x| - \frac{1}{2} \log |1 - x| - \frac{1}{2} \log |1 + x| + c$. $I = \frac{1}{2} [2 \log |x| - \log |1 - x| - \log |1 + x|] + c$. $I = \frac{1}{2} \log \left| \frac{x^2}{(1 - x)(1 + x)} \right| + c = \frac{1}{2} \log \frac{x^2}{1 - x^2} + c$.

$\int \frac{1}{x - x^3} \, dx = $

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