int frac{x , dx}{x^2 + 4x + 5} = | vedclass

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(D) Let $I = \int \frac{x \, dx}{x^2 + 4x + 5}$.We can rewrite the numerator as $x = \frac{1}{2}(2x + 4) - 2 = \frac{1}{2}(2x + 4) - 2$.So,$I = \int \

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$\frac{1}{2}\log(x^2 + 4x + 5) - 2	an^{-1}(x + 2) + c$

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(D) Let $I = \int \frac{x \, dx}{x^2 + 4x + 5}$.We can rewrite the numerator as $x = \frac{1}{2}(2x + 4) - 2 = \frac{1}{2}(2x + 4) - 2$.So,$I = \int \frac{\frac{1}{2}(2x + 4) - 2}{x^2 + 4x + 5} \, dx$.$I = \frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 5} \, dx - 2 \int \frac{dx}{(x+2)^2 + 1}$.For the first integral,let $t = x^2 + 4x + 5$,then $dt = (2x + 4) \, dx$.$I = \frac{1}{2} \int \frac{dt}{t} - 2 \int \frac{dx}{(x+2)^2 + 1^2}$.Using the standard integrals $\int \frac{1}{t} \, dt = \log|t|$ and $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{2} \log|x^2 + 4x + 5| - 2 	an^{-1}(x + 2) + c$.

$\int \frac{x \, dx}{x^2 + 4x + 5} = $

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