Calculate the improper integral: integral from 0 to 1

Calculate the improper integral: $$ \int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx $$

This is an improper integral because the integrand has a vertical asymptote at $x = 1$.

Step 1: Identify the singularity

As $x \to 1^-$:

$$x - 1 \to 0^- \implies (x-1)^{2/3} \to 0^+ \implies \frac{x+1}{(x-1)^{2/3}} \to +\infty$$

Note: $(x-1)^{2/3} = [(x-1)^2]^{1/3}$ is always positive for $x \neq 1$.

Step 2: Express as a limit

$$\int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx = \lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{x+1}{(x-1)^{2/3}} \, dx$$

Step 3: Use substitution

Let $u = x - 1$, so $x = u + 1$ and $dx = du$.

When $x = 0$: $u = -1$

When $x = 1 - \varepsilon$: $u = -\varepsilon$

The integral becomes:

$$\int_0^{1-\varepsilon} \frac{x+1}{(x-1)^{2/3}} \, dx = \int_{-1}^{-\varepsilon} \frac{(u+1)+1}{u^{2/3}} \, du = \int_{-1}^{-\varepsilon} \frac{u+2}{u^{2/3}} \, du$$

Step 4: Split the integral

$$\int_{-1}^{-\varepsilon} \frac{u+2}{u^{2/3}} \, du = \int_{-1}^{-\varepsilon} \frac{u}{u^{2/3}} \, du + \int_{-1}^{-\varepsilon} \frac{2}{u^{2/3}} \, du$$

$$= \int_{-1}^{-\varepsilon} u^{1/3} \, du + 2\int_{-1}^{-\varepsilon} u^{-2/3} \, du$$

Step 5: Evaluate the first integral

$$\int_{-1}^{-\varepsilon} u^{1/3} \, du = \left[\frac{u^{4/3}}{4/3}\right]_{-1}^{-\varepsilon} = \frac{3}{4}\left[u^{4/3}\right]_{-1}^{-\varepsilon}$$

For $u < 0$, we have $u^{4/3} = (u^4)^{1/3} = |u|^{4/3}$ (since $u^4 > 0$).

More carefully, for $u = -a$ where $a > 0$:

$$u^{4/3} = (-a)^{4/3} = [(-a)^4]^{1/3} = (a^4)^{1/3} = a^{4/3}$$

So:

$$(-\varepsilon)^{4/3} = \varepsilon^{4/3} \quad \text{and} \quad (-1)^{4/3} = 1$$

Therefore:

$$\int_{-1}^{-\varepsilon} u^{1/3} \, du = \frac{3}{4}(\varepsilon^{4/3} - 1)$$

Step 6: Evaluate the second integral

$$\int_{-1}^{-\varepsilon} u^{-2/3} \, du = \left[\frac{u^{1/3}}{1/3}\right]_{-1}^{-\varepsilon} = 3[u^{1/3}]_{-1}^{-\varepsilon}$$

For $u = -a$ where $a > 0$:

$$u^{1/3} = (-a)^{1/3} = -a^{1/3}$$

So:

$$(-\varepsilon)^{1/3} = -\varepsilon^{1/3} \quad \text{and} \quad (-1)^{1/3} = -1$$

Therefore:

$$\int_{-1}^{-\varepsilon} u^{-2/3} \, du = 3[(-\varepsilon)^{1/3} - (-1)^{1/3}] = 3[-\varepsilon^{1/3} - (-1)] = 3(1 - \varepsilon^{1/3})$$

Step 7: Combine the results

$$\int_0^{1-\varepsilon} \frac{x+1}{(x-1)^{2/3}} \, dx = \frac{3}{4}(\varepsilon^{4/3} - 1) + 2 \cdot 3(1 - \varepsilon^{1/3})$$

$$= \frac{3}{4}\varepsilon^{4/3} - \frac{3}{4} + 6 - 6\varepsilon^{1/3}$$

$$= \frac{3}{4}\varepsilon^{4/3} - 6\varepsilon^{1/3} + 6 - \frac{3}{4}$$

$$= \frac{3}{4}\varepsilon^{4/3} - 6\varepsilon^{1/3} + \frac{21}{4}$$

Step 8: Take the limit

$$\int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx = \lim_{\varepsilon \to 0^+} \left(\frac{3}{4}\varepsilon^{4/3} - 6\varepsilon^{1/3} + \frac{21}{4}\right)$$

As $\varepsilon \to 0^+$:

$$\varepsilon^{4/3} \to 0 \quad \text{and} \quad \varepsilon^{1/3} \to 0$$

Therefore:

$$\int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx = 0 - 0 + \frac{21}{4} = \frac{21}{4}$$

Final Answer

$$\boxed{\int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx = \frac{21}{4}}$$

The integral converges to $\frac{21}{4} = 5.25$.

Why the integral converges

Near $x = 1$, we can analyze the behavior:

$$\frac{x+1}{(x-1)^{2/3}} \approx \frac{2}{(x-1)^{2/3}}$$

This is of the form $\frac{1}{(1-x)^{2/3}}$, which corresponds to:

$$\int_a^1 \frac{dx}{(1-x)^{\alpha}} \quad \text{with } \alpha = \frac{2}{3}$$

This integral converges if and only if $\alpha < 1$.

Since $\frac{2}{3} < 1$, the integral converges! ✓

Verification of convergence criterion

The general result states:

$$\int_a^b \frac{dx}{(b-x)^\alpha} \text{ converges} \iff \alpha < 1$$

In our case, the dominant term near $x = 1$ is $\frac{2}{(x-1)^{2/3}}$, which has $\alpha = \frac{2}{3} < 1$.

Therefore, convergence is guaranteed.

Alternative approach: Direct integration

We can also integrate directly without the limit:

$$\int \frac{x+1}{(x-1)^{2/3}} \, dx$$

Using the substitution $u = x - 1$:

$$= \int \frac{u+2}{u^{2/3}} \, du = \int u^{1/3} \, du + 2\int u^{-2/3} \, du$$

$$= \frac{3u^{4/3}}{4} + 2 \cdot 3u^{1/3} + C = \frac{3u^{4/3}}{4} + 6u^{1/3} + C$$

$$= \frac{3(x-1)^{4/3}}{4} + 6(x-1)^{1/3} + C$$

Evaluating from 0 to 1:

$$\left[\frac{3(x-1)^{4/3}}{4} + 6(x-1)^{1/3}\right]_0^1$$

At $x = 1$: both terms are 0

At $x = 0$:

$$\frac{3(-1)^{4/3}}{4} + 6(-1)^{1/3} = \frac{3 \cdot 1}{4} + 6(-1) = \frac{3}{4} - 6 = -\frac{21}{4}$$

Therefore:

$$0 - \left(-\frac{21}{4}\right) = \frac{21}{4}$$

This confirms our answer! ✓

Numerical verification

Let's check by computing $\int_0^{0.9} \frac{x+1}{(x-1)^{2/3}} \, dx$:

At $x = 0.9$:

$$\frac{3(-0.1)^{4/3}}{4} + 6(-0.1)^{1/3} \approx \frac{3(0.0464)}{4} + 6(-0.464) \approx 0.0348 - 2.784 \approx -2.749$$

At $x = 0$: $-\frac{21}{4} = -5.25$

Difference: $-2.749 - (-5.25) = 2.501 \approx 2.5$

As we integrate closer to $x = 1$, we approach $\frac{21}{4} = 5.25$.

Summary

1. The integral has a singularity at $x = 1$ (upper endpoint)
2. Using the substitution $u = x - 1$ transforms it to standard forms
3. The integral splits into: $$ \int u^{1/3} \, du + 2\int u^{-2/3} \, du $$
4. The exponent $\frac{2}{3} < 1$ ensures convergence
5. Both integrals can be evaluated using the power rule
6. The final result is: $$ \int_0^1 \frac{x+1}{(x-1)^{2/3}} \, dx = \frac{21}{4} $$

Key lessons:

• Substitution $u = x - b$ is useful for singularities at $x = b$
• Fractional exponents require careful handling with negative values
• The convergence criterion $\alpha < 1$ is fundamental
• Splitting complex integrands into simpler parts aids computation
• Verification through alternative methods confirms results
• For $u < 0$: $u^{p/q}$ depends on whether $p$ is even or odd