The value of int frac{dx}{3 - 2x - x^2} is | vedclass

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(A) We need to evaluate the integral $I = \int \frac{dx}{3 - 2x - x^2}$.First,complete the square in the denominator: $3 - 2x - x^2 = 4 - (x^2 + 2x +

Vedclass · Accepted Answer

$\frac{1}{4} \log \left( \frac{3 + x}{1 - x} \right)$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int \frac{dx}{3 - 2x - x^2}$.First,complete the square in the denominator: $3 - 2x - x^2 = 4 - (x^2 + 2x + 1) = 4 - (x + 1)^2$.So,$I = \int \frac{dx}{4 - (x + 1)^2}$.Let $t = x + 1$,then $dt = dx$.Substituting this into the integral,we get $I = \int \frac{dt}{2^2 - t^2}$.Using the standard formula $\int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \log \left| \frac{a + t}{a - t} \right| + C$,where $a = 2$:$I = \frac{1}{2(2)} \log \left| \frac{2 + t}{2 - t} \right| + C = \frac{1}{4} \log \left| \frac{2 + (x + 1)}{2 - (x + 1)} \right| + C = \frac{1}{4} \log \left| \frac{3 + x}{1 - x} \right| + C$.Thus,the correct option is $A$.

The value of $\int \frac{dx}{3 - 2x - x^2}$ is

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