allery$\int \dfrac{dx}{x^2-6x+13}$ equals
A$\dfrac{1}{2}\tan^{-1}\dfrac{x-3}{2}+C$
B$\dfrac{1}{2}\log\left|\dfrac{x-3}{x+3}\right|+C$
C$\tan^{-1}(x-3)+C$
D$\dfrac{1}{4}\tan^{-1}\dfrac{x-3}{2}+C$
Answer & Solution
Correct answer: A. $\dfrac{1}{2}\tan^{-1}\dfrac{x-3}{2}+C$
1. Complete the square: $x^2-6x+13=(x-3)^2+4=(x-3)^2+2^2$.
2. Put $t=x-3$, $dt=dx$.
3. Integral $=\int \dfrac{dt}{t^2+2^2}=\dfrac{1}{2}\tan^{-1}\dfrac{t}{2}$.
4. Back-substitute: $\dfrac{1}{2}\tan^{-1}\dfrac{x-3}{2}+C$. Option D has the wrong $\tfrac{1}{a}$ factor.
_Source: syllabus Mathematics Ch 7 "Integrals", p.310_
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