Evaluate the integral: I = integral x/((x²-3x+2)

Evaluate the integral: $$ I = \int \frac{x}{\left(x^2-3x+2\right)\sqrt{x^2-4x+3}}\,dx. $$

Step 1 · Factor and shift

With $y=x-2$: $x^2-3x+2=y(y+1)$, $x^2-4x+3=y^2-1$, $x=y+2$:

$$ I = \int\frac{(y+2)\,dy}{y(y+1)\sqrt{y^2-1}}. $$

Step 2 · Trig substitution

Let $y=\sec\varphi$ (for $y>1$, i.e. $x>3$), so $dy=\sec\varphi\tan\varphi\,d\varphi$ and $\sqrt{y^2-1}=\tan\varphi$:

$$ I = \int\frac{(\sec\varphi+2)\sec\varphi\tan\varphi\,d\varphi}{\sec\varphi(\sec\varphi+1)\tan\varphi} = \int\frac{\sec\varphi+2}{\sec\varphi+1}\,d\varphi. $$

Step 3 · Integrate

Writing $\dfrac{\sec\varphi+2}{\sec\varphi+1}=1+\dfrac1{\sec\varphi+1}=1+\dfrac{\cos\varphi}{1+\cos\varphi}$, and using $1+\cos\varphi=2\cos^2(\varphi/2)$:

$$ \frac{\cos\varphi}{1+\cos\varphi} = 1 - \frac1{2\cos^2(\varphi/2)}. $$

$$ \int\frac{\sec\varphi+2}{\sec\varphi+1}\,d\varphi = \int\left[2-\frac1{2\cos^2(\varphi/2)}\right]d\varphi = 2\varphi - \tan\left(\frac\varphi2\right) + C. $$

Step 4 · Back-substitute twice

With $y=\sec\varphi$: $\varphi=\arccos(1/y)$, and $\tan(\varphi/2)=\dfrac{\sqrt{y^2-1}}{y+1}$:

$$ I = 2\arccos\left(\frac1y\right) - \frac{\sqrt{y^2-1}}{y+1} + C, $$

and with $y=x-2$, $y^2-1=x^2-4x+3$, $y+1=x-1$:

Summary Table

Additional Notes

• Major rewrite: the source's Step 5 abandoned its own correct substitution setup ("this is quite complex... let's try a different approach") and asserted a final answer, $-1/\sqrt{x^2-4x+3}$, purely by pattern-matching, without deriving it. That answer fails a differentiation check at $x=5$ ($-0.133$ vs. the true integrand $0.147$). The boxed result below completes the trig-substitution approach the source itself set up correctly in Steps 1–3 and was verified independently.
• The domain (matching the source's own test points) is $x>3$.
• The result was verified by differentiation at $x=4,5,6$: the derivative matches the integrand $x/[(x^2-3x+2)\sqrt{x^2-4x+3}]$ at all three points (e.g. $\approx0.3849$ at $x=4$).