If \(\int\frac{\left(2x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1}dx=C-\frac{1}{f\left(x\right)}\), where \(f\left(x\right)\) is of the form of \(ax^{2}+bx+c\) then a+b+c is equal to

\(5\)

Let, \(I= \int{\frac{(2x + 3)dx}{x(x + 1)(x + 2)(x + 3) +1}}\)

            \(\int{\frac{(2x + 3)dx}{(x^2 + 3x)(x^2 + 3x + 2) + 1}}\)

Putting, \(x^2 +3x = t\)

             \(⇒ (2x + 3)dx = dt\)

\(∴ I =  \int{\frac{dt}{t(t + 2) + 1}}\)

            \(\int{(t + 1)^2}\)

           \(= c − \frac{1}{(t + 1)}\)

           \(= c − \frac{1}{x^2 + 3x + 1}\)

On comparing with \(c −\frac{1}{ax^2 + bx + c}\), we get

\(a = 1; b = 3 and c = 1 \)

\(∴ a + b + c = 5\)