$\int \frac{d x}{(x+2) \sqrt{x+1}}=$

$\int \frac{d x}{(x+2) \sqrt{x+1}}=$
  1. $\tan ^{-1}(\sqrt{x+1})+c$
  2. $2 \tan ^{-1}(\sqrt{x+1})+c$
  3. $2 \tan ^{-1}(\sqrt{x+2})+c$
  4. $\tan ^{-1}(\sqrt{x+2})+c$

Solution

Let $I=\int \frac{d x}{(x+2) \sqrt{x+1}}$ Put $\sqrt{x+1}=t \Rightarrow(x+1)=t^{2}$ and $d x=2 t d t$ $\therefore I=\int \frac{2 t d t}{\left(t^{2}+1\right) t}$ $\quad=2 \int \frac{d t}{t^{2}+1}=2 \tan ^{-1} t+c=2 \tan ^{-1}(\sqrt{x+1})+c$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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