$\int(x+1)(x+2)^4(x+3) d x$ is equal to

$\int(x+1)(x+2)^4(x+3) d x$ is equal to
  1. $\frac{(x+1)^2}{2}+\frac{(x+2)^2}{5}+\frac{(x+3)^2}{2}+C$
  2. $\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C$
  3. $\frac{(x+2)^7}{7}+\frac{(x+2)^5}{5}+C$
  4. $\frac{(x+3)^7}{7}-\frac{(x+3)^5}{5}+C$

Solution

Let $x+2=t$, then $d x=d t$ $ \begin{aligned} I & =\int(t-1)(t)^4(t+1) d t=\int\left(t^2-1\right) t^4 d t \\ & =\int\left(t^6-t^4\right) d t \\ & =\frac{t^7}{7}-\frac{t^5}{5}+C \\ & =\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C \end{aligned} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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