By completing the square,
t^2-6t+13=(t^2-6t+9)+4=(t-3)^2+2^2t2−6t+13=(t2−6t+9)+4=(t−3)2+22
So, we can rewrite the integral as
int{dt}/{sqrt{(t-3)^2+2^2}}∫dt√(t−3)2+22
Let t-3=2tan thetat−3=2tanθ.
Rightarrow {dt}/{d theta}=2sec^2 theta
Rightarrow dt=2sec^2 theta d theta⇒dtdθ=2sec2θ⇒dt=2sec2θdθ
by the above substitution,
=int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)}
=int{sec^2theta}/{sqrt{tan^2theta+1}}d theta=∫2sec2θdθ√(2tanθ)2+22=∫sec2θ√tan2θ+1dθ
by the identity tan^2theta+1=sec^2thetatan2θ+1=sec2θ,
=int{sec^2theta}/{sqrt{sec^2theta}}d theta=int sec theta d theta=ln|sec theta + tan theta|+C_1=∫sec2θ√sec2θdθ=∫secθdθ=ln|secθ+tanθ|+C1
by using
tan theta={t-3}/2tanθ=t−32 and sec theta=sqrt{tan^2theta+1}={sqrt{t^2-6+13}}/2secθ=√tan2θ+1=√t2−6+132,
we have
=ln|{sqrt{t^2-6+13}}/2+{t-3}/2|+C_1=ln∣∣∣√t2−6+132+t−32∣∣∣+C1
=ln|{sqrt{t^2-6t+13}+t-3}/2|+C_1=ln∣∣∣√t2−6t+13+t−32∣∣∣+C1
by log property,
=ln|sqrt{t^2-6t+13}+t-3|-ln2+C_1=ln∣∣√t2−6t+13+t−3∣∣−ln2+C1
by setting C=ln2+C_1C=ln2+C1,
=ln|sqrt{t^2-6t+13}+t-3|+C=ln∣∣√t2−6t+13+t−3∣∣+C
