Complete the square before choosing a form

Write $t^2+4t+13=(t+2)^2-4+13=(t+2)^2+9=(t+2)^2+3^2$. With $f(t)=t+2$, $f^{\prime}(t)=1$, the integral is $\frac13\tan^{-1}\bigl(\frac{t+2}{3}\bigr)+C$. The Teacher suggests comparing coefficients in $(t+a)^2+b$: $2a=4$, $a=2$, $a^2+b=13$, $b=9$.

$2x^2+8x+9=2(x^2+4x+9/2)=2[(x+2)^2-4+9/2]=2[(x+2)^2+1/2]$. Therefore the integral is $\frac12\int\frac1{(x+2)^2+(1/\sqrt2)^2}\,dx=\frac12\frac1{1/\sqrt2}\tan^{-1}\bigl(\frac{x+2}{1/\sqrt2}\bigr)+C=\frac{\sqrt2}{2}\tan^{-1}[\sqrt2(x+2)]+C$.

$13+8x-4x^2=13-4(x^2-2x)=13-4[(x-1)^2-1]=17-4(x-1)^2$. Taking $\sqrt4=2$ outside the radical gives $\frac12\int\frac1{\sqrt{17/4-(x-1)^2}}\,dx=\frac12\sin^{-1}\bigl(\frac{x-1}{\sqrt{17}/2}\bigr)+C=\frac12\sin^{-1}\bigl[\frac2{\sqrt{17}}(x-1)\bigr]+C$, for $|x-1|<\sqrt{17}/2$.