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