Evaluate the integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dt}{t^2+10t+41}$

$ta{n}^{-1}(t-5)+C$

$-5t+C$

$2ta{n}^{-1}(\frac{t+5}{4})+C$

$\frac{1}{4}ta{n}^{-1}(\frac{t+5}{4})+C$