If int frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = fr | vedclass

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(C) We have,$\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$. Rearranging the denominator: $x(x+3)(x+1)(x+2)+1 = (x^2+3x)(x^2

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$5$

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(C) We have,$\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$. Rearranging the denominator: $x(x+3)(x+1)(x+2)+1 = (x^2+3x)(x^2+3x+2)+1$. Let $t = x^2+3x$,then $dt = (2x+3)dx$. Substituting into the integral: $\int \frac{dt}{t(t+2)+1} = \int \frac{dt}{t^2+2t+1} = \int \frac{dt}{(t+1)^2}$. Integrating gives: $-\frac{1}{t+1} + \alpha = \frac{-1}{x^2+3x+1} + \alpha$. Comparing this with $\frac{-1}{ax^2+bx+c} + \alpha$,we get $a=1, b=3, c=1$. Therefore,$a+b+c = 1+3+1 = 5$.

If $\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$,then the value of $a+b+c$ is equal to

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