Let's solve the two integrals step by step:

(a) $\int_0^1 \frac{1}{(x-1)^{4/5}} \, dx$

The integral has a singularity at $x = 1$, where the denominator becomes zero. To evaluate this, we can rewrite the integral and analyze whether it converges or diverges.

We are dealing with: $\int_0^1 (x-1)^{-4/5} \, dx$ Shifting the variable $u = x - 1$, so that when $x = 0$, $u = -1$, and when $x = 1$, $u = 0$. Thus, the integral becomes: $\int_{-1}^0 u^{-4/5} \, du$ This is an improper integral near $u = 0$, which is a singularity point. The general form of this type of integral is: $\int u^{p} \, du$ where $p = -4/5$. If $p > -1$, the integral converges. Since $p = -4/5 > -1$, the integral converges.

Solving: $\int u^{-4/5} \, du = \frac{u^{1/5}}{1/5} = 5u^{1/5}$ Evaluating this from $u = -1$ to $u = 0$: $5 \left[ 0^{1/5} - (-1)^{1/5} \right]$ Since $(-1)^{1/5} = -1$, the result is: $5(0 - (-1)) = 5$ Thus, the value of the integral is: $\boxed{5}$

(b) $\int_1^\infty \frac{\ln x}{x^3} \, dx$

This is another improper integral, but this time over an infinite range. We can use integration by parts to solve it.

We rewrite the integral as: $\int_1^\infty \ln x \cdot x^{-3} \, dx$ Using integration by parts, let:

• $u = \ln x$, so $du = \frac{1}{x} \, dx$
• $dv = x^{-3} \, dx$, so $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$

Now apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$ Substituting: $\left[ -\frac{\ln x}{2x^2} \right]_1^\infty - \int_1^\infty -\frac{1}{2x^2} \cdot \frac{1}{x} \, dx$ The first term evaluates as: $\lim_{x \to \infty} -\frac{\ln x}{2x^2} + \frac{\ln 1}{2} = 0 + 0 = 0$ For the second term: $\int_1^\infty \frac{1}{2x^3} \, dx = \frac{1}{4x^2} \Big|_1^\infty = \frac{1}{4}$ Thus, the value of the second integral is: $\boxed{\frac{1}{4}}$

Summary:

• (a) The value of the first integral is $\boxed{5}$.
• (b) The value of the second integral is $\boxed{\frac{1}{4}}$.

Would you like further details on any part of the solution?

Here are five related questions for further exploration:

1. How do you handle improper integrals with singularities?
2. What is the method of substitution, and when is it useful in integration?
3. What conditions determine the convergence of an improper integral?
4. How does integration by parts simplify the evaluation of integrals involving logarithms?
5. What are the applications of logarithmic integrals in real-world problems?

Tip: When handling improper integrals, always check for convergence at both bounds before attempting to solve the integral.