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…

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
  1. 3
  2. 4
  3. 5
  4. 6

Solution

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 $x^2+3 x=t \Rightarrow(2 x+3) d x=d t$ $ \begin{aligned} & \therefore \text { LHS }=\int \frac{d t}{t^2+2 t+1} \\ &=\int \frac{d t}{(t+1)^2}=-\frac{1}{t+1}+\alpha \\ &=\frac{-1}{x^2+3 x+1}+\alpha \\ & a=1, b=3, c=1 \\ & \therefore \quad a+b+c=1+3+1=5 \end{aligned} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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