Indefinite integration of a fraction with a non-factorable denominator

Solve the integral below:

$$\int \frac{x+1}{x^2-4x+6} \, dx$$

I tried u-sub and got

$$u=x^2-4x+6$$ $$du = 2x - 4 dx \leftrightarrow dx = \frac{1}{2(x-2)}du$$ $$\int \frac{x+1}{u} \frac{1}{2(x-2)} \, du = \frac{1}{2}\int \frac{x+1}{x-2} \frac{1}{u} \, du$$

Since that didn't lead me anywhere, I tried long division, which didn't help either. I was considering partial fraction decomposition however, I can't factor the denominator, so I'm stuck. Where do I go from here? Please let me know if any further clarification is necessary.

You can write the integral as$$\frac 12\int\frac{2x-4}{x^2-4x+6}dx+\int\frac{3}{(x-2)^2+2}dx$$ $$=\frac 12\ln(x^2-4x+6)+\frac{3}{\sqrt{2}}\arctan\left(\frac{x-2}{\sqrt{2}}\right)+c$$
What you want to do is complete the square in the denominator, and the apply the appropriate trigonometric substitution which in this case will be $x-2=\sqrt2\tan\theta$: \begin{align}\int\frac{x+1}{x^2-4x+6}dx&=\int\frac{(x-2)+3}{(x-2)^2+2}dx=\int\frac{\sqrt2\tan\theta+3}{2\sec^2\theta}\sqrt2\sec^2\theta\,d\theta\\ &=\int\left(\tan\theta+\frac3{\sqrt2}\right)d\theta=-\ln\cos\theta+\frac3{\sqrt2}\theta+C_1\\ &=-\ln\left(\frac{\sqrt2}{\sqrt{(x-2)^2+2}}\right)+\frac3{\sqrt2}\tan^{-1}\frac{x-2}{\sqrt2}+C_1\\ &=\frac12\ln(x^2-4x+6)+\frac3{\sqrt2}\tan^{-1}\frac{x-2}{\sqrt2}+C\end{align} Differentiation confirms this result.
• Completing the square means trying to write $x^2+bx+c$ somehow as $\left(x+\frac b2\right)^2+d$. If you expand the latter expression out, it reads $x^2+bx+\frac{b^2}4+d=x^2+bx+c$. So it's a perfect match already, except we need $\frac{b^2}4+d=c$, so we find that $d=c-\frac{b^2}4$. In this case $b=-4$ and $c=6$ so we conclude that $d=6-\frac{(-4)^2}4=2$ and $x^2-4x+6=\left(x+\frac{(-4)}2\right)^2+2=\left(x-2\right)^2+2$. May 20 '16 at 11:22