text{If } int frac{3x+2}{4x^2+4x+5} dx = A lo | vedclass

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Question

(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$. We express the numerator as a derivative of the denominator: $3x+2 = \frac{3}{8}(8x+4) + (2 - \frac{12}{

Vedclass · Accepted Answer

$(\frac{3}{8}, \frac{1}{8})$

Vedclass · Answer

(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$. We express the numerator as a derivative of the denominator: $3x+2 = \frac{3}{8}(8x+4) + (2 - \frac{12}{8}) = \frac{3}{8}(8x+4) + \frac{1}{2}$. Substituting this into the integral: $I = \int \frac{\frac{3}{8}(8x+4) + \frac{1}{2}}{4x^2+4x+5} dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5} dx + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$. For the first part,let $u = 4x^2+4x+5$,then $du = (8x+4)dx$. $I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$. Using the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} 	an^{-1}(\frac{2x+1}{2}) + C$. $I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{8} 	an^{-1}(x+\frac{1}{2}) + C$. Comparing with the given form,$A = \frac{3}{8}$ and $B = \frac{1}{8}$.

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

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