The value of $∫\frac{dx}{x^2-6x+13}$ is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{1}{2}\tan^{-1}\frac{x-3}{2}+C, $ where C is constant of integration.
$∫\frac{dx}{x^2-6x+13}$
$=∫\frac{dx}{(x-3)^2+2^2}$
$=\frac{1}{2}\tan^{-1}\frac{(x+3)}{2}+C$