Find the following integral: int frac{x+2}{2 | vedclass

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Question

(A) To evaluate $\int \frac{x+2}{2 x^{2}+6 x+5} d x$,we express the numerator as $A \frac{d}{d x}(2 x^{2}+6 x+5) + B$.$x+2 = A(4x+6) + B = 4Ax + (6A+B

Vedclass · Accepted Answer

$\frac{1}{4} \log |2 x^{2}+6 x+5|+\frac{1}{2} 	an ^{-1}(2 x+3)+ C$

Vedclass · Answer

(A) To evaluate $\int \frac{x+2}{2 x^{2}+6 x+5} d x$,we express the numerator as $A \frac{d}{d x}(2 x^{2}+6 x+5) + B$.$x+2 = A(4x+6) + B = 4Ax + (6A+B)$.Equating coefficients,$4A = 1 \implies A = \frac{1}{4}$ and $6A + B = 2 \implies 6(\frac{1}{4}) + B = 2 \implies B = 2 - \frac{3}{2} = \frac{1}{2}$.Thus,the integral becomes $\frac{1}{4} \int \frac{4x+6}{2x^2+6x+5} dx + \frac{1}{2} \int \frac{1}{2x^2+6x+5} dx$.Let $I_1 = \int \frac{4x+6}{2x^2+6x+5} dx = \log |2x^2+6x+5| + C_1$.For $I_2 = \int \frac{1}{2x^2+6x+5} dx = \frac{1}{2} \int \frac{1}{x^2+3x+\frac{5}{2}} dx = \frac{1}{2} \int \frac{1}{(x+\frac{3}{2})^2 + (\frac{1}{2})^2} dx$.Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$,we get $I_2 = \frac{1}{2} \cdot \frac{1}{1/2} 	an^{-1}(\frac{x+3/2}{1/2}) + C_2 = 	an^{-1}(2x+3) + C_2$.Combining these,the final result is $\frac{1}{4} \log |2 x^{2}+6 x+5|+\frac{1}{2} 	an ^{-1}(2 x+3)+ C$.

Find the following integral: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$

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