int frac{x+1}{x^{2}+5 x+6} d x= | vedclass

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Question

(A) Let $I = \int \frac{x+1}{x^{2}+5x+6} dx$. Factorizing the denominator: $x^{2}+5x+6 = (x+2)(x+3)$. Using partial fractions: $\frac{x+1}{(x+2)(x+3)}

Vedclass · Accepted Answer

$-\log |x+2|+2 \log |x+3|+C$

Vedclass · Answer

(A) Let $I = \int \frac{x+1}{x^{2}+5x+6} dx$. Factorizing the denominator: $x^{2}+5x+6 = (x+2)(x+3)$. Using partial fractions: $\frac{x+1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$. $x+1 = A(x+3) + B(x+2)$. Setting $x = -2$: $-2+1 = A(-2+3) \implies A = -1$. Setting $x = -3$: $-3+1 = B(-3+2) \implies -2 = B(-1) \implies B = 2$. Thus,$I = \int \left( \frac{-1}{x+2} + \frac{2}{x+3} \right) dx$. $I = -\int \frac{1}{x+2} dx + 2 \int \frac{1}{x+3} dx$. $I = -\log |x+2| + 2 \log |x+3| + C$.

$\int \frac{x+1}{x^{2}+5 x+6} d x=$

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