int frac{dx}{x - x^2} = | vedclass

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Question

(A) To evaluate the integral $I = \int \frac{dx}{x - x^2}$,we first factor the denominator: $x - x^2 = x(1 - x)$. Using partial fraction decomposition

Vedclass · Accepted Answer

$\log |x| - \log |1 - x| + C$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{dx}{x - x^2}$,we first factor the denominator: $x - x^2 = x(1 - x)$. Using partial fraction decomposition,we write: $\frac{1}{x(1 - x)} = \frac{A}{x} + \frac{B}{1 - x}$. Multiplying by $x(1 - x)$,we get $1 = A(1 - x) + Bx$. Setting $x = 0$,we find $A = 1$. Setting $x = 1$,we find $B = 1$. Thus,$\frac{1}{x(1 - x)} = \frac{1}{x} + \frac{1}{1 - x}$. Integrating both sides with respect to $x$: $I = \int \left( \frac{1}{x} + \frac{1}{1 - x} \right) dx$. $I = \log |x| - \log |1 - x| + C$ (since the derivative of $1 - x$ is $-1$). Therefore,the correct option is $A$.

$\int \frac{dx}{x - x^2} = $

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