int frac{d x}{x^{2}+4 x+13} = | vedclass

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Question

(A) To evaluate the integral $\int \frac{d x}{x^{2}+4 x+13}$,we complete the square in the denominator: $x^{2}+4 x+13 = (x^{2}+4 x+4) + 9 = (x+2)^{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To evaluate the integral $\int \frac{d x}{x^{2}+4 x+13}$,we complete the square in the denominator: $x^{2}+4 x+13 = (x^{2}+4 x+4) + 9 = (x+2)^{2} + 3^{2}$. Now,the integral becomes $\int \frac{d x}{(x+2)^{2} + 3^{2}}$. Using the standard formula $\int \frac{d x}{x^{2}+a^{2}} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight) + c$,where $a=3$ and the variable is $(x+2)$: $\int \frac{d x}{(x+2)^{2} + 3^{2}} = \frac{1}{3} 	an^{-1}\left(\frac{x+2}{3}ight) + c$. Comparing this with the given options,the correct option is $A$.

$\int \frac{d x}{x^{2}+4 x+13} = $

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