int frac{dx}{x^2 + 4x + 13} is equal to | vedclass

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Question

(B) To evaluate the integral $I = \int \frac{dx}{x^2 + 4x + 13}$,we first complete the square in the denominator.$x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 =

Vedclass · Accepted Answer

$\frac{1}{3} 	an^{-1} \left( \frac{x + 2}{3} ight) + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{x^2 + 4x + 13}$,we first complete the square in the denominator.$x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 3^2$.Now,the integral becomes $I = \int \frac{dx}{(x + 2)^2 + 3^2}$.Using the standard integration formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) + c$,where $a = 3$ and the variable is $(x + 2)$:$I = \frac{1}{3} 	an^{-1} \left( \frac{x + 2}{3} ight) + c$.Thus,the correct option is $B$.

$\int \frac{dx}{x^2 + 4x + 13}$ is equal to

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