If int frac{x+5}{x^2+4x+5} dx = a log(x^2+4x+ | vedclass

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(A) Let $I = \int \frac{x+5}{x^2+4x+5} dx$. We express the numerator as $x+5 = \lambda(2x+4) + \mu$. Comparing the coefficients of $x$ and the constan

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$(\frac{1}{2}, 3, 2)$

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(A) Let $I = \int \frac{x+5}{x^2+4x+5} dx$. We express the numerator as $x+5 = \lambda(2x+4) + \mu$. Comparing the coefficients of $x$ and the constant terms on both sides: $1 = 2\lambda \implies \lambda = \frac{1}{2}$ $5 = 4\lambda + \mu \implies 5 = 4(\frac{1}{2}) + \mu \implies 5 = 2 + \mu \implies \mu = 3$. Thus, $I = \int \frac{\frac{1}{2}(2x+4) + 3}{x^2+4x+5} dx = \frac{1}{2} \int \frac{2x+4}{x^2+4x+5} dx + 3 \int \frac{1}{(x+2)^2 + 1^2} dx$. Using the formula $\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$ and $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = \frac{1}{2} \log(x^2+4x+5) + 3 	an^{-1}(x+2) + C$. Comparing this with the given form $a \log(x^2+4x+5) + b 	an^{-1}(x+k) + C$, we get $a = \frac{1}{2}$, $b = 3$, and $k = 2$.

If $\int \frac{x+5}{x^2+4x+5} dx = a \log(x^2+4x+5) + b \tan^{-1}(x+k) + C$, then $(a, b, k)$ equals

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