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TS EAMCET · Maths · Indefinite Integration

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 \(\frac{3 p-q}{r}=\)

  1. A 0
  2. B 1
  3. C 2
  4. D -1
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(A) 0

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We have, \[ \begin{aligned} I & =\int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x \\ & =-\frac{1}{p x^2+q x+r}+c \\ I & =\int \frac{2 x+3}{\left(x^2+3 x\right)\left(x^2+3 x+2\right)+1} d x \end{aligned} \] Put \(x^2+3 x=t \Rightarrow(2 x+3) d x=d t\)…
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