If int {frac{{left( {2x + 3} right)dx}}{{xlef | vedclass

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(B) Let $I = \int \frac{(2x + 3)dx}{x(x + 1)(x + 2)(x + 3) + 1}$.Rearrange the denominator: $[x(x + 3)][(x + 1)(x + 2)] + 1 = (x^2 + 3x)(x^2 + 3x + 2)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $I = \int \frac{(2x + 3)dx}{x(x + 1)(x + 2)(x + 3) + 1}$.Rearrange the denominator: $[x(x + 3)][(x + 1)(x + 2)] + 1 = (x^2 + 3x)(x^2 + 3x + 2) + 1$.Let $t = x^2 + 3x$. Then $dt = (2x + 3)dx$.The integral becomes $I = \int \frac{dt}{t(t + 2) + 1} = \int \frac{dt}{t^2 + 2t + 1} = \int \frac{dt}{(t + 1)^2}$.Integrating,we get $I = -\frac{1}{t + 1} + C = C - \frac{1}{x^2 + 3x + 1}$.Comparing this with $C - \frac{1}{f(x)}$,we have $f(x) = x^2 + 3x + 1$.Thus,$a = 1, b = 3, c = 1$.Therefore,$a + b + c = 1 + 3 + 1 = 5$.

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

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