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Magazine Chapter 7: Problem 50
\(I=\int \frac{x^{4}-1}{x^{2} \cdot x \sqrt{x^{2}+\frac{1}{x^{2}}+1}} d x=\int \frac{\left(x-\frac{1}{x^{3}}\right) d x}{\sqrt{x^{2}+\frac{1}{x^{2}}+1}}\) Put \(x^{2}+\frac{1}{x^{2}}+1=t\) or \(2\left(x-\frac{1}{x^{3}}\right) d x=d t\)
Short Answer
Expert verified
The integral is \( 2\sqrt{x^{2} + \frac{1}{x^{2}} + 1} + C \).
Step by step solution
01
Understand the Transformation
The given integral is \( I = \int \frac{x^{4} - 1}{x^{2} \cdot x \sqrt{x^{2} + \frac{1}{x^{2}} + 1}} \, dx = \int \frac{\left(x-\frac{1}{x^{3}}\right) \, dx}{\sqrt{x^{2} + \frac{1}{x^{2}} + 1}} \). We choose a substitution to simplify it.
02
Choose a Substitution
We let \( t = x^{2} + \frac{1}{x^{2}} + 1 \). This implies that \( 2\left(x - \frac{1}{x^{3}}\right) \, dx = dt \).
03
Differentiate the Substitution
Differentiate both sides of the substitution with respect to \( x \): \( d(x^{2}) = 2x \, dx \) and \( d(\frac{1}{x^{2}}) = -\frac{2}{x^{3}} \, dx \), so the differential becomes \( dt = 2\left(x - \frac{1}{x^{3}}\right) \, dx \).
04
Plug in the Substitution
Substitute \( t \) and \( dt \) into the integral: \[ I = \int \frac{2\left(x - \frac{1}{x^{3}}\right)}{\sqrt{t}} \, \frac{dt}{2\left(x - \frac{1}{x^{3}}\right)} = \int \frac{1}{\sqrt{t}} \, dt \]
05
Integrate with Respect to t
The integral is now simpler: \( I = \int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt \). The antiderivative is \( 2t^{1/2} + C \).
06
Substitute Back for x
Substitute back \( t = x^{2} + \frac{1}{x^{2}} + 1 \) so that the expression in terms of \( x \) is \( 2\sqrt{x^{2} + \frac{1}{x^{2}} + 1} + C \).
07
Final Result
The solution to the original integral is \( I = 2\sqrt{x^{2} + \frac{1}{x^{2}} + 1} + C \), where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in integration that simplifies the process of finding an integral. In this approach, we replace complex expressions with a single variable to make integration easier. For instance, in the exercise, we chose to set \(t = x^2 + \frac{1}{x^2} + 1\). This substitution allows us to transform the original integral into a simpler form.
How does this work?
How does this work?
- We identify a part of the integral that can be set as a new variable \(t\).
- We differentiate this new variable with respect to \(x\) to find \(dt\).
- Finally, we substitute both \(t\) and \(dt\) into the integral, simplifying it considerably.
Definite Integral
A definite integral has specific upper and lower limits and computes the signed area under a curve. Unlike indefinite integrals, definite integrals result in a number, not a function. Although the solution provided focuses on indefinite integration, understanding definite integrals is crucial.
Here鈥檚 how it works:
Here鈥檚 how it works:
- The integral is evaluated between two bounds, say \(a\) and \(b\).
- We use the antiderivative to compute the integral from \(a\) to \(b\).
- The result gives the net area under the function between these bounds, accounting for areas above and below the x-axis.
Antiderivative
An antiderivative is a function whose derivative is the original function you are working with. It's a crucial component in integration because it helps to reverse differentiation.
Steps to find an antiderivative:
Steps to find an antiderivative:
- Write the integrand (function to integrate) in a more manageable form if needed.
- Find the antiderivative, sometimes using rules such as power rule, trigonometric rules, or exponential rules.
- Include a constant of integration, symbolized by \(C\), to represent the family of solutions.
Constant of Integration
The constant of integration, denoted as \(C\), is a constant added to the antiderivative of a function. It represents all the possible vertical shifts of the antiderivative on a graph.
Why is it important?
Why is it important?
- When differentiating a constant, its derivative is zero, so any number added originally would vanish. Thus, integrating again, we introduce \(C\) to account for all potential values.
- This constant ensures that we represent the entire family of antiderivatives, not just one particular solution.
- In definite integrals, \(C\) cancels out when evaluated between the limits, which is why it does not appear in the final result.
