$\int \frac{1}{x^2+4 x+13} d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$I =\int \frac{1}{x^2+4 x+13} d x$
$=\int \frac{1}{(x+2)^2+3^2} d x=\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)+C$
$\int \frac{1}{x^2+4 x+13} d x$ is equal to
Correct answer: option 2
We have,
$I =\int \frac{1}{x^2+4 x+13} d x$
$=\int \frac{1}{(x+2)^2+3^2} d x=\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)+C$