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Antiderivative

If ∫2x +3dxxx...

Question

If $\int \frac{(2 x + 3) d x}{x (x + 1) (x + 2) (x + 3) + 1} = K - \frac{1}{f (x)}$ where $f (x)$ is of the form of $a x^{2} + b x + c$ and $K$ is the constant of integration, then $(a + b + c)$ equals

A

2

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B

5

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C

9

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D

13

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Solution

The correct option is B $5$
$I = \int \frac{(2 x + 3) d x}{x (x + 1) (x + 2) (x + 3) + 1} I = \int \frac{(2 x + 3) d x}{(x^{2} + 3 x + 2) (x^{2} + 3 x) + 1}$
Put $x^{2} + 3 x = t \Rightarrow (2 x + 3) d x = d t$

$I = \int \frac{d t}{t (t + 2) + 1} = \int \frac{d t}{(t + 1)^{2}} = K - \frac{1}{t + 1} = K - \frac{1}{x^{2} + 3 x + 1} ∴ a = 1, b = 3, c = 1 \Rightarrow a + b + c = 5$

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