int_0^1 frac{2 x+5}{x^2+3 x+2} ,d x= | vedclass

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(A) First, factor the denominator: $x^2+3 x+2 = (x+1)(x+2)$.Using partial fractions, let $\frac{2 x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.Eq

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$\log \left(\frac{16}{3}\right)$

Vedclass · Answer

(A) First, factor the denominator: $x^2+3 x+2 = (x+1)(x+2)$.Using partial fractions, let $\frac{2 x+5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.Equating numerators: $2 x+5 = A(x+2) + B(x+1)$.For $x = -1$, $2(-1)+5 = A(-1+2) \implies A = 3$.For $x = -2$, $2(-2)+5 = B(-2+1) \implies 1 = -B \implies B = -1$.So, $\int_0^1 \left(\frac{3}{x+1} - \frac{1}{x+2}\right) \,d x = [3 \log |x+1| - \log |x+2|]_0^1$.Evaluating at limits: $(3 \log 2 - \log 3) - (3 \log 1 - \log 2) = 3 \log 2 - \log 3 + \log 2 = 4 \log 2 - \log 3 = \log(16) - \log(3) = \log \left(\frac{16}{3}\right)$.

$\int_0^1 \frac{2 x+5}{x^2+3 x+2} \,d x=$

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