Problem 40
Question
Find the indefinite integral. $$\int u^{-2}\left(1-u^{2}+u^{4}\right) d u$$
Step-by-Step Solution
Verified Answer
The indefinite integral of the given function is:
\[
\int u^{-2}(1-u^2+u^4)du = - \frac{1}{u} - u + \frac{1}{3}u^3 + C
\]
1Step 1: Expand the function and simplify
To make the integration process simpler, first, expand the function by multiplying each term in the parentheses with the \(u^{-2}\) term. This will result in three separate terms that we can integrate individually.
\[
\int u^{-2}(1-u^2+u^4)du = \int (u^{-2} - u^0 + u^2) du
\]
2Step 2: Integrate each term
Now that we have three separate terms, we can integrate each of them individually. For each term, raise the exponent by 1 and then divide by the new exponent.
\[
\int (u^{-2}-u^0+u^2) du = \frac{u^{-1}}{-1} - \frac{u^1}{1} + \frac{u^3}{3} + C
\]
Where C is the constant of integration.
3Step 3: Simplify and write the final answer
Finally, let's simplify the expression and write down the final answer.
\[
- \frac{1}{u} - u + \frac{1}{3}u^3 + C
\]
So, the indefinite integral of the given function is:
\[
\int u^{-2}(1-u^2+u^4)du = - \frac{1}{u} - u + \frac{1}{3}u^3 + C
\]
Key Concepts
Polynomial ExpansionIntegration TechniquesConstant of Integration
Polynomial Expansion
Polynomial expansion is a technique that simplifies expressions, especially when dealing with integration. In our exercise, we began with an integral that contained a polynomial inside parentheses combined with a power of a variable. This can be a bit tricky at first.
The polynomial we started with was \(1 - u^2 + u^4\), and it was multiplied by \(u^{-2}\). To expand this, we distribute the \(u^{-2}\) across each term in the polynomial, resulting in three separate terms: \(u^{-2} \cdot 1 = u^{-2}\), \(u^{-2} \cdot (-u^2) = -u^0\), and \(u^{-2} \cdot u^4 = u^2\). This simplifies our integral to \(\int (u^{-2} - 1 + u^2) du\). This step makes integration much easier.
The polynomial we started with was \(1 - u^2 + u^4\), and it was multiplied by \(u^{-2}\). To expand this, we distribute the \(u^{-2}\) across each term in the polynomial, resulting in three separate terms: \(u^{-2} \cdot 1 = u^{-2}\), \(u^{-2} \cdot (-u^2) = -u^0\), and \(u^{-2} \cdot u^4 = u^2\). This simplifies our integral to \(\int (u^{-2} - 1 + u^2) du\). This step makes integration much easier.
- Key Point: Expand by multiplying each term separately.
- Why it's useful: Simplifies complex functions for integration.
- Resultant Terms: Each separate term \(u^{-2}, -1, u^2\) is easier to integrate.
Integration Techniques
Integration can seem challenging, but by breaking down the process, it's a lot more approachable. With the expanded polynomial \(u^{-2} - 1 + u^2\), we integrate each term individually, which is the essence of many integration techniques.
To integrate powers of \(u\), the general rule is to increase the exponent by one and divide by the new exponent:
To integrate powers of \(u\), the general rule is to increase the exponent by one and divide by the new exponent:
- For \(\int u^{-2} du\), increasing the exponent from \(-2\) to \(-1\) gives \(\frac{u^{-1}}{-1}\).
- For \(\int (-1) du\), since \(-1\) is a constant, its integral is simply \(-u\).
- For \(\int u^2 du\), increasing the exponent to \(3\) gives \(\frac{u^3}{3}\).
Constant of Integration
The constant of integration \(C\) is a crucial part of indefinite integration. Whenever you find an indefinite integral, there are infinitely many antiderivatives for a function. Each antiderivative can differ by a constant.
Including the constant \(C\) accounts for all these possibilities, ensuring your solution is complete. In our exercise, after integrating the terms individually, we add \(C\) to reflect this uncertainty:
Including the constant \(C\) accounts for all these possibilities, ensuring your solution is complete. In our exercise, after integrating the terms individually, we add \(C\) to reflect this uncertainty:
- The integral of \(u^{-2}\) gives \(-\frac{1}{u}\).
- Integrating \(-1\) results in \(-u\).
- Integrating \(u^2\) results in \(\frac{1}{3}u^3\).
- Adding \(C\) accounts for the infinite antiderivatives.
Other exercises in this chapter
Problem 40
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