int frac{dx}{1 - x^2} = | vedclass

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(C) To evaluate the integral $\int \frac{dx}{1 - x^2}$,we use the method of partial fractions.The integrand can be written as $\frac{1}{1 - x^2} = \fr

Vedclass · Accepted Answer

$\frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| + c$

Vedclass · Answer

(C) To evaluate the integral $\int \frac{dx}{1 - x^2}$,we use the method of partial fractions.The integrand can be written as $\frac{1}{1 - x^2} = \frac{1}{(1 - x)(1 + x)}$.Using partial fractions,we have $\frac{1}{(1 - x)(1 + x)} = \frac{A}{1 - x} + \frac{B}{1 + x}$.Solving for $A$ and $B$,we get $1 = A(1 + x) + B(1 - x)$.Setting $x = 1$,we get $1 = 2A \implies A = \frac{1}{2}$.Setting $x = -1$,we get $1 = 2B \implies B = \frac{1}{2}$.Thus,$\int \frac{dx}{1 - x^2} = \int \left( \frac{1/2}{1 - x} + \frac{1/2}{1 + x} \right) dx$.Integrating term by term,we get $\frac{1}{2} \int \frac{dx}{1 - x} + \frac{1}{2} \int \frac{dx}{1 + x} = -\frac{1}{2} \ln |1 - x| + \frac{1}{2} \ln |1 + x| + c$.Using the property of logarithms,$\ln a - \ln b = \ln(a/b)$,we get $\frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| + c$.

$\int \frac{dx}{1 - x^2} = $

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