If int frac{(2 x+3)}{x(x+1)(x+2)(x+3)+1} d x | vedclass

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(A) We have, $I = \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x$. Rearranging the terms in the denominator: $x(x+3) = x^2+3x$ and $(x+1)(x+2) = x^2+3x+2$.

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(A) We have, $I = \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x$. Rearranging the terms in the denominator: $x(x+3) = x^2+3x$ and $(x+1)(x+2) = x^2+3x+2$. So, $I = \int \frac{2 x+3}{(x^2+3 x)(x^2+3 x+2)+1} d x$. Let $t = x^2+3 x$, then $dt = (2x+3) dx$. Substituting these into the integral: $I = \int \frac{d t}{t(t+2)+1} = \int \frac{d t}{t^2+2 t+1} = \int \frac{d t}{(t+1)^2}$. Integrating, we get $I = -\frac{1}{t+1} + c$. Substituting $t = x^2+3x$ back, we get $I = -\frac{1}{x^2+3 x+1} + c$. Comparing this with $-\frac{1}{p x^2+q x+r} + c$, we find $p=1, q=3, r=1$. Therefore, $\frac{3 p-q}{r} = \frac{3(1)-3}{1} = \frac{0}{1} = 0$.

If $\int \frac{(2 x+3)}{x(x+1)(x+2)(x+3)+1} d x =-\frac{1}{p x^2+q x+r}+c$, then $\frac{3 p-q}{r}=$

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