Calculate the improper integral: integral from 0 to ²

Calculate the improper integral: $$ \int_0^2 \frac{dx}{x^2-3x+2} $$

This integral requires careful analysis because the denominator may have zeros within the interval of integration.

Step 1: Factor the denominator

First, let's factor $x^2 - 3x + 2$:

$$x^2 - 3x + 2 = (x-1)(x-2)$$

We can verify: $(x-1)(x-2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$ ✓

Step 2: Identify singularities

The denominator equals zero when:

$$(x-1)(x-2) = 0 \implies x = 1 \text{ or } x = 2$$

Both $x = 1$ and $x = 2$ are in the interval $[0, 2]$!

• $x = 1$ is an interior singularity
• $x = 2$ is a singularity at the right endpoint

The integrand has vertical asymptotes at both points.

Step 3: Split the integral at the singularity

Since we have a singularity at $x = 1$ in the interior, we must split:

$$\int_0^2 \frac{dx}{x^2-3x+2} = \int_0^1 \frac{dx}{(x-1)(x-2)} + \int_1^2 \frac{dx}{(x-1)(x-2)}$$

Both integrals are now improper and must be evaluated as limits.

Step 4: Analyze the sign of the integrand

For $x \in [0, 1)$:

• $x - 1 < 0$ (negative)
• $x - 2 < 0$ (negative)
• $(x-1)(x-2) > 0$ (product of two negatives is positive)

For $x \in (1, 2)$:

• $x - 1 > 0$ (positive)
• $x - 2 < 0$ (negative)
• $(x-1)(x-2) < 0$ (product is negative)

So $\frac{1}{(x-1)(x-2)}$ is positive on $[0, 1)$ and negative on $(1, 2)$.

Step 5: Use partial fractions

$$\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$$

Multiplying both sides by $(x-1)(x-2)$:

$$1 = A(x-2) + B(x-1)$$

Setting $x = 1$: $1 = A(-1) \implies A = -1$

Setting $x = 2$: $1 = B(1) \implies B = 1$

Therefore:

$$\frac{1}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{1}{x-2} = \frac{1}{x-2} - \frac{1}{x-1}$$

Step 6: Evaluate the first integral $\int_0^1 \frac{dx}{(x-1)(x-2)}$

$$\int_0^1 \frac{dx}{(x-1)(x-2)} = \int_0^1 \left(\frac{1}{x-2} - \frac{1}{x-1}\right) dx$$

$$= \lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \left(\frac{1}{x-2} - \frac{1}{x-1}\right) dx$$

$$= \lim_{\varepsilon \to 0^+} \left[\ln|x-2| - \ln|x-1|\right]_0^{1-\varepsilon}$$

$$= \lim_{\varepsilon \to 0^+} \left[\ln\left|\frac{x-2}{x-1}\right|\right]_0^{1-\varepsilon}$$

At $x = 1 - \varepsilon$:

$$\ln\left|\frac{(1-\varepsilon)-2}{(1-\varepsilon)-1}\right| = \ln\left|\frac{-1-\varepsilon}{-\varepsilon}\right| = \ln\left(\frac{1+\varepsilon}{\varepsilon}\right) = \ln(1+\varepsilon) - \ln\varepsilon$$

As $\varepsilon \to 0^+$:

$$\ln(1+\varepsilon) \to 0 \quad \text{but} \quad \ln\varepsilon \to -\infty$$

So: $\ln(1+\varepsilon) - \ln\varepsilon \to 0 - (-\infty) = +\infty$

Therefore:

$$\int_0^1 \frac{dx}{(x-1)(x-2)} = +\infty \quad \text{(diverges)}$$

Conclusion

Since the first integral diverges, the entire integral diverges.

Final Answer

$$\boxed{\int_0^2 \frac{dx}{x^2-3x+2} = +\infty \quad \text{(diverges)}}$$

The integral diverges due to the singularity at $x = 1$.

Why the integral diverges

Near $x = 1$, let's analyze the behavior. For $x$ slightly less than 1:

$$\frac{1}{(x-1)(x-2)} \approx \frac{1}{(x-1)(-1)} = \frac{-1}{x-1} = \frac{1}{1-x}$$

This behaves like $\frac{1}{1-x}$ near $x = 1$, which corresponds to:

$$\int_a^1 \frac{dx}{1-x}$$

This is the integral $\int \frac{dx}{(1-x)^1}$ with exponent $\alpha = 1$.

Since $\alpha = 1 \geq 1$, the integral diverges! This is the critical case of logarithmic divergence.

More detailed analysis near $x = 1$

From the left (as $x \to 1^-$):

$$\int_{1-\delta}^{1-\varepsilon} \frac{dx}{(x-1)(x-2)} \approx \int_{1-\delta}^{1-\varepsilon} \frac{dx}{1-x}$$

$$= [-\ln|1-x|]_{1-\delta}^{1-\varepsilon} = -\ln\varepsilon + \ln\delta = \ln\frac{\delta}{\varepsilon}$$

As $\varepsilon \to 0$, this goes to infinity.

Verification using partial fractions directly

Let's verify the divergence more explicitly:

$$\int_0^{1-\varepsilon} \left(\frac{1}{x-2} - \frac{1}{x-1}\right) dx$$

The first term:

$$\int_0^{1-\varepsilon} \frac{dx}{x-2} = [\ln|x-2|]_0^{1-\varepsilon} = \ln(2-1+\varepsilon) - \ln(2) = \ln(1+\varepsilon) - \ln 2$$

As $\varepsilon \to 0^+$: this approaches $\ln 1 - \ln 2 = -\ln 2$ (finite)

The second term:

$$-\int_0^{1-\varepsilon} \frac{dx}{x-1} = -[\ln|x-1|]_0^{1-\varepsilon} = -\ln|1-\varepsilon-1| + \ln|0-1|$$

$$= -\ln\varepsilon + \ln 1 = -\ln\varepsilon$$

As $\varepsilon \to 0^+$: $-\ln\varepsilon \to +\infty$

Therefore, the integral diverges.

What if we only integrated to a value less than 1?

If we consider $\int_0^a \frac{dx}{x^2-3x+2}$ for $0 < a < 1$, this would be a proper integral (no singularities) and would converge.

For example:

$$\int_0^{1/2} \frac{dx}{(x-1)(x-2)} = \left[\ln\left|\frac{x-2}{x-1}\right|\right]_0^{1/2}$$

$$= \ln\left|\frac{-3/2}{-1/2}\right| - \ln\left|\frac{-2}{-1}\right| = \ln 3 - \ln 2 = \ln\frac{3}{2}$$

This is finite.

Graphical interpretation

The function $f(x) = \frac{1}{x^2-3x+2} = \frac{1}{(x-1)(x-2)}$ has:

• Vertical asymptote at $x = 1$ (goes to $+\infty$ from left, $-\infty$ from right)
• Vertical asymptote at $x = 2$ (goes to $-\infty$ from left, $+\infty$ from right)
• The function is positive on $(-\infty, 1) \cup (2, \infty)$
• The function is negative on $(1, 2)$

The area under the curve from 0 to 1 is infinite due to the singularity at $x = 1$.

Summary

1. The integral has singularities at $x = 1$ (interior) and $x = 2$ (endpoint)
2. We factor: $x^2 - 3x + 2 = (x-1)(x-2)$
3. Using partial fractions: $\frac{1}{(x-1)(x-2)} = \frac{1}{x-2} - \frac{1}{x-1}$
4. Near $x = 1$, the integrand behaves like $\frac{1}{1-x}$
5. This gives a logarithmic singularity: $\int \frac{dx}{1-x} = -\ln|1-x|$
6. The integral diverges because of the $\alpha = 1$ critical exponent
7. Result: $\int_0^2 \frac{dx}{x^2-3x+2} = +\infty$

Key lessons:

• Always check for zeros of the denominator in the integration interval
• Interior singularities require splitting the integral
• Partial fractions are essential for rational functions
• The critical exponent $\alpha = 1$ causes logarithmic divergence
• Simple poles (linear factors in denominator) lead to logarithmic singularities
• Not all improper integrals converge - this one diverges