AP EAMCET · Maths · Indefinite Integration
If \(\int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x\) \(=\frac{-1}{a x^2+b x+c}+\alpha\), then value of \(a+b+c\) is equal to
- A 3
- B 4
- C 5
- D 6
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x=\frac{-1}{a x^2+b x+c}+\alpha \\ & \text { LHS }=\int \frac{2 x+3}{x(x+3)(x+1)(x+2)+1} d x \\ & \quad=\int \frac{(2 x+3) d x}{\left(x^2+3 x\right)\left(x^2+3 x+2\right)+1} \end{aligned} \] Put…
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