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If $\int \frac{d x}{x^2+2 x+2}=f(x)+c$, then $f(x)$ is equal to :
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Verified Answer
The correct answer is:
$\tan ^{-1}(x+1)$
Let $I=\int \frac{d x}{x^2+2 x+2}$
$=\int \frac{d x}{x^2+2 x+1+1}=\int \frac{d x}{1+(x+1)^2}$
$=\tan ^{-1}(x+1)+c$
But $I=f(x)+c$
$\therefore \quad f(x)=\tan ^{-1}(x+1)$
$=\int \frac{d x}{x^2+2 x+1+1}=\int \frac{d x}{1+(x+1)^2}$
$=\tan ^{-1}(x+1)+c$
But $I=f(x)+c$
$\therefore \quad f(x)=\tan ^{-1}(x+1)$
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