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$\begin{aligned} & \text { If } \int \frac{3 x+2}{4 x^2+4 x+5} d x=A \log \left(4 x^2+4 x+5\right)+B \operatorname{Tan}^{-1} \\ & \left(x+\frac{1}{2}\right)+C, \text { then }(A, B)=\end{aligned}$
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Verified Answer
The correct answer is:
$\left(\frac{3}{8}, \frac{1}{8}\right)$
Let $I=\int \frac{3 x+2}{4 x^2+4 x+5} d x$
$\begin{aligned} & =\frac{3}{8} \int \frac{8 x+4}{4 x^2+4 x+5} d x+\frac{1}{2} \int \frac{1}{4 x^2+4 x+5} d x \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \int \frac{1}{(2 x+1)^2+2^2} d x \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \times \frac{1}{4} \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{8} \tan ^{-1}\left(x+\frac{1}{2}\right)+C\end{aligned}$
$\begin{aligned} & =\frac{3}{8} \int \frac{8 x+4}{4 x^2+4 x+5} d x+\frac{1}{2} \int \frac{1}{4 x^2+4 x+5} d x \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \int \frac{1}{(2 x+1)^2+2^2} d x \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \times \frac{1}{4} \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{8} \tan ^{-1}\left(x+\frac{1}{2}\right)+C\end{aligned}$
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