Evaluate Integral Of Ln(s)/s^6 Using Integration By Parts

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Hey guys! Let's dive into a fun mathematical problem today: evaluating the definite integral $\int_1^2 \frac{\ln(s)}{s^6} ds$. This might look a bit intimidating at first, but don't worry! We're going to break it down step by step using a technique called integration by parts. This method is super useful when you have a product of two functions inside an integral, like we do here with $\ln(s)$ and $\frac{1}{s^6}$. So, grab your thinking caps, and let's get started!

Before we jump into the problem itself, let's quickly recap what integration by parts is all about. The formula for integration by parts comes directly from the product rule for differentiation. Remember that? If we have two functions, $u(s)$ and $v(s)$, the product rule tells us:

$\frac{d}{ds}[u(s)v(s)] = u'(s)v(s) + u(s)v'(s)$

Now, if we integrate both sides of this equation with respect to $s$, we get:

$\int \frac{d}{ds}[u(s)v(s)] ds = \int [u'(s)v(s) + u(s)v'(s)] ds$

The left side simply becomes $u(s)v(s)$, and we can split the integral on the right side:

$u(s)v(s) = \int u'(s)v(s) ds + \int u(s)v'(s) ds$

Rearranging this equation gives us the integration by parts formula:

$\int u(s)v'(s) ds = u(s)v(s) - \int v(s)u'(s) ds$

This formula is the key to solving our integral! The idea is to choose our $u$ and $v'$ such that the new integral on the right-hand side is simpler to evaluate than the original one. This often involves strategically picking a $u$ that becomes simpler when differentiated. When facing these types of problems, identifying appropriate parts is crucial, and the acronym LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) can serve as a valuable guide. LIATE helps you prioritize which function to designate as u in the integration by parts formula. Following this order typically makes the integration process smoother. In our case, $\ln(s)$ is a logarithmic function and $\frac{1}{s^6}$ is an algebraic function. According to LIATE, we should choose $\ln(s)$ as our u. This choice is strategic because the derivative of $\ln(s)$ is $\frac{1}{s}$, which simplifies the expression, potentially making the integral easier to handle. On the other hand, if we chose $\frac{1}{s^6}$ as u, we would have to differentiate it, resulting in $-6s^{-7}$, which does not simplify the integral as much.

Okay, let's get back to our integral: $\int_1^2 \frac{\ln(s)}{s^6} ds$. We need to choose our $u$ and $v'$. As we discussed, a good choice here is:

• $u = \ln(s)$
• $v' = \frac{1}{s^6} = s^{-6}$

Now we need to find $u'$ and $v$. Differentiating $u$ is straightforward:

$u' = \frac{1}{s}$

To find $v$, we need to integrate $v'$:

$v = \int s^{-6} ds = \frac{s^{-5}}{-5} = -\frac{1}{5s^5}$

Now we have everything we need to apply the integration by parts formula:

$\int u(s)v'(s) ds = u(s)v(s) - \int v(s)u'(s) ds$

Plugging in our values:

$\int \frac{\ln(s)}{s^6} ds = \ln(s)\left(-\frac{1}{5s^5}\right) - \int \left(-\frac{1}{5s^5}\right)\left(\frac{1}{s}\right) ds$

Let's simplify this a bit:

$\int \frac{\ln(s)}{s^6} ds = -\frac{\ln(s)}{5s^5} + \frac{1}{5} \int \frac{1}{s^6} ds$

Look at that! The integral on the right is much simpler than our original one. We've made progress!

Now we just need to evaluate $\int \frac{1}{s^6} ds$. We already did this when we found $v$, but let's do it again for clarity:

$\int \frac{1}{s^6} ds = \int s^{-6} ds = \frac{s^{-5}}{-5} = -\frac{1}{5s^5}$

So, our integral becomes:

$\int \frac{\ln(s)}{s^6} ds = -\frac{\ln(s)}{5s^5} + \frac{1}{5} \left(-\frac{1}{5s^5}\right) + C$

Simplifying further:

$\int \frac{\ln(s)}{s^6} ds = -\frac{\ln(s)}{5s^5} - \frac{1}{25s^5} + C$

We've found the indefinite integral, but we need to evaluate the definite integral from 1 to 2:

$\int_1^2 \frac{\ln(s)}{s^6} ds = \left[-\frac{\ln(s)}{5s^5} - \frac{1}{25s^5}\right]_1^2$

This means we need to plug in our limits of integration:

$\left[-\frac{\ln(2)}{5(2)^5} - \frac{1}{25(2)^5}\right] - \left[-\frac{\ln(1)}{5(1)^5} - \frac{1}{25(1)^5}\right]$

Remember that $\ln(1) = 0$, so this simplifies to:

$\left[-\frac{\ln(2)}{5(32)} - \frac{1}{25(32)}\right] - \left[0 - \frac{1}{25}\right]$

$\left[-\frac{\ln(2)}{160} - \frac{1}{800}\right] + \frac{1}{25}$

To make things easier, let's find a common denominator, which is 800:

$\left[-\frac{5\ln(2)}{800} - \frac{1}{800}\right] + \frac{32}{800}$

Combining the terms:

$\frac{-5\ln(2) - 1 + 32}{800} = \frac{31 - 5\ln(2)}{800}$

So, our final answer is:

$\int_1^2 \frac{\ln(s)}{s^6} ds = \frac{31 - 5\ln(2)}{800}$

Awesome! We successfully evaluated the integral $\int_1^2 \frac{\ln(s)}{s^6} ds$ using integration by parts. It might have seemed a bit tricky at first, but by breaking it down into smaller steps and using the integration by parts formula, we were able to find the solution. Remember, the key to mastering integration by parts is practice, practice, practice! The integration by parts formula is not just a mathematical tool; it's a method that showcases the power of strategic thinking in problem-solving. The goal is to transform a difficult integral into something more manageable, and this involves selecting the parts (u and v') wisely. In our example, the strategic choice of $u = \ln(s)$ simplified the integral by reducing the logarithmic function to a simple reciprocal after differentiation. This strategy reflects a core principle in mathematics: simplifying complex problems by transforming them into simpler, solvable forms. Moreover, the integration by parts technique highlights the fundamental relationship between differentiation and integration, illustrating how reversing the product rule can lead to elegant solutions. The process not only yields a result but also provides a deeper appreciation for the interconnectedness of mathematical operations. The result $\frac{31 - 5\ln(2)}{800}$ is more than just a numerical answer; it encapsulates a journey through mathematical techniques, strategic decisions, and the beauty of mathematical relationships. It embodies the essence of calculus, where complex problems are systematically broken down and resolved through thoughtful application of fundamental principles. Keep practicing, and you'll become an integration ninja in no time!

• Integration by parts
• Definite integral
• Logarithmic function
• Algebraic function
• LIATE rule
• Calculus
• Integral evaluation
• Step-by-step solution
• Mathematical techniques
• Problem-solving