If $\int \frac{1}{x^2+2 x+2} d x=f(x)+C$, then $f(x)=$
Answer & explanation
Correct answer: option 1
We have,
$\int \frac{1}{x^2+2 x+2} d x=\int \frac{1}{(x+1)^2+1^2} d x=\tan ^{-1}(x+1)+C$
$f(x)=\tan ^{-1}(x+1)$
If $\int \frac{1}{x^2+2 x+2} d x=f(x)+C$, then $f(x)=$
Correct answer: option 1
We have,
$\int \frac{1}{x^2+2 x+2} d x=\int \frac{1}{(x+1)^2+1^2} d x=\tan ^{-1}(x+1)+C$
$f(x)=\tan ^{-1}(x+1)$