int frac{3x+2}{4x^2+4x+5} dx = A log(4x^2+4x+ | vedclass

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(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$. To solve this, we express the numerator as a derivative of the denominator: $3x+2 = k \frac{d}{dx}(4x^2+

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$. To solve this, we express the numerator as a derivative of the denominator: $3x+2 = k \frac{d}{dx}(4x^2+4x+5) + m = k(8x+4) + m = 8kx + (4k+m)$. Comparing coefficients: $8k = 3 \implies k = \frac{3}{8}$ and $4k+m = 2 \implies 4(\frac{3}{8}) + m = 2 \implies \frac{3}{2} + m = 2 \implies m = \frac{1}{2}$. Thus, $I = \int \frac{\frac{3}{8}(8x+4) + \frac{1}{2}}{4x^2+4x+5} dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5} dx + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$. $I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \cdot \frac{1}{2} 	an^{-1}(\frac{2x+1}{2}) \cdot \frac{1}{2} + c$ (adjusting for the coefficient of $x$). Actually, $\int \frac{1}{(2x+1)^2 + 2^2} dx = \frac{1}{2} 	an^{-1}(\frac{2x+1}{2}) \cdot \frac{1}{2} = \frac{1}{4} 	an^{-1}(\frac{2x+1}{2})$. So, $I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \cdot \frac{1}{4} 	an^{-1}(\frac{2x+1}{2}) + c = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{8} 	an^{-1}(\frac{2x+1}{2}) + c$. Comparing with the given form, $A = \frac{3}{8}$ and $B = \frac{1}{8}$. Therefore, $A+B = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

$\int \frac{3x+2}{4x^2+4x+5} dx = A \log(4x^2+4x+5) + B \tan^{-1}\left(\frac{2x+1}{2}\right) + c$, then $A+B=$

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