Integrals Involving Quadratics — Question 5

Question 5 – Solution

Step 1: Complete the square. $$\begin{align*} x^2-6x+10 &=(x^2-6x+9)+1\\ &=(x-3)^2+1. \end{align*}$$ Step 2: Substitute $u=x-3$. Then $du=dx$ and $$I=\int\frac{du}{\sqrt{u^2+1}}.$$ Step 3: Use the standard formula. $$\int\frac{du}{\sqrt{u^2+a^2}} =\ln\left|u+\sqrt{u^2+a^2}\right|+C.$$ With $a=1$, $$\begin{align*} I&=\ln\left|u+\sqrt{u^2+1}\right|+C\\ &=\ln\left|x-3+\sqrt{x^2-6x+10}\right|+C. \end{align*}$$ $$\boxed{\ln|x-3+\sqrt{x^2-6x+10}|+C}$$