Indefinite integral: $\int \frac{\sqrt{x^2-6x+18}}{x-3}dx$

Question

I have the following indefinite at hand and I'm sure substitution is the only way I should go about solving this, but each time I think I get close, I end up at the same place, which doesn't seem to be the solution according to WolframAlpha.

$$\int \frac{\sqrt{x^2-6x+18}}{x-3}dx$$

I tried to substitute with $t=x-3$ for the denominator and a step after that $u=t^2+9$ for the numerator under the square root, but I would like to hand it over to someone who could perhaps show me a way!

After you have made this substitution, you can let $t=\tan\theta$ and $dt=\sec^{2}\theta d\theta$. — user84413, Aug 22 '14 at 18:51
Also you can use Euler's first substitution: $$\sqrt{x^2+6x+18}=x+t$$ — pointer, Aug 22 '14 at 21:25

score 1 · Accepted Answer · answered Aug 22 '14 at 19:05

After you make the standard substitution $t=3\tan\theta$, you end up needing to integrate something which apart from constants looks like $\frac{\sec^3\theta}{\tan\theta}$. What next?

There are several possibilities, none very pleasant.

One way is to rewrite the integrand as $\frac{\sec^3\theta\tan\theta}{\tan^2\theta}$, and then as $\frac{\sec^3\theta\tan\theta}{\sec^2\theta-1}$. Then the substitution $w=\sec\theta$ leaves us with $\int\frac{w^2}{w^2-1}$, which is relatively easy (after division, partial fractions).

Some other options are to write $\frac{\sec^{3}\theta}{\tan\theta}$ as $\frac{\sec\theta}{\tan\theta}(\tan^{2}\theta+1)$ and simplify, or as $(\csc\theta)(\sec^{2}\theta)$ and then integrate by parts. — user84413, Aug 22 '14 at 19:22

score 0 · Answer 2 · answered Feb 25 '23 at 19:50

Substitute $t= \sqrt{x^2-6x+18}>3$

\begin{align} &\int \frac{\sqrt{x^2-6x+18}}{x-3}\ dx\\ =&\ \frac12\int \left( 1-\frac{9}{9-t^2}\right)dt = \frac12t-\frac32\coth^{-1}\frac t3\\ =& \ \frac12 \sqrt{x^2-6x+18}-\frac32\coth^{-1}\frac{\sqrt{x^2-6x+18}}3 \end{align}

Indefinite integral: $\int \frac{\sqrt{x^2-6x+18}}{x-3}dx$

2 Answers2