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Chapter 25: Problem 34

Integrate each of the given expressions. \(\int \frac{2 x^{2} d x}{\sqrt{2 x^{3}+1}}\)

Short Answer

Expert verified
The integral is \( \frac{2}{3} \sqrt{2x^3 + 1} + C \).

Step by step solution

01

Identify the substitution

To simplify the given integral \( \int \frac{2x^2 \, dx}{\sqrt{2x^3 + 1}} \), notice that the expression \(2x^3 + 1\) is inside the square root. A smart substitution would be setting \(u = 2x^3 + 1\), meaning the derivative of \(u\) concerning \(x\) is needed.
02

Perform substitution

Differentiate \(u = 2x^3 + 1\) with respect to \(x\): \( \frac{du}{dx} = 6x^2 \). Therefore, \( du = 6x^2 \, dx \). To substitute in the integral, notice that \(x^2 \, dx\) can be expressed as \( \frac{1}{6} du \) due to \( du = 6x^2 \, dx \). Now substitute in the integral: \( \int \frac{2x^2 \, dx}{\sqrt{2x^3 + 1}} = \int \frac{2}{6} \cdot \frac{du}{\sqrt{u}} = \int \frac{1}{3} \cdot u^{-1/2} \, du \).
03

Integrate with respect to \(u\)

Perform the integral \( \int \frac{1}{3} u^{-1/2} \, du \). The power rule for integration says \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) if \(n eq -1\). Here \(n = -1/2\), so we have: \[ \int u^{-1/2} \, du = u^{1/2} + C. \] Therefore, \( \frac{1}{3} \int u^{-1/2} \, du = \frac{1}{3} \left( 2u^{1/2} \right) + C = \frac{2}{3} u^{1/2} + C \).
04

Substitute back

Replace \(u\) with the original expression in terms of \(x\): \( u = 2x^3 + 1 \). So, the integrated result is \[ \frac{2}{3} (2x^3 + 1)^{1/2} + C. \]
05

Express the final answer

The result of the integral \( \int \frac{2x^2 \, dx}{\sqrt{2x^3 + 1}} \) is \[ \frac{2}{3} \sqrt{2x^3 + 1} + C, \] where \(C\) is a 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
When you see a complex integral, such as \( \int \frac{2x^2 \, dx}{\sqrt{2x^3 + 1}} \), you might feel overwhelmed. This is where the Substitution Method comes to the rescue! This technique simplifies your integral by changing variables, essentially transforming it into an easier problem.
  • First, identify a part of the expression to substitute. Often, it's the inside of a composite function, like \(2x^3 + 1\) inside the square root.
  • Next, replace this with a new variable, say \(u\). In this case, let \(u = 2x^3 + 1\).
  • This means you'll need the derivative \(\frac{du}{dx}\), which here is \(6x^2\). Rearrange \(du = 6x^2 \, dx\) and use it to transform the integral fully into terms of \(u\).
Now, your integral is much simpler: \(\int \frac{1}{3} u^{-1/2} \, du\). It's a bit like putting together a puzzle; once you have the right pieces, it fits perfectly.
Power Rule for Integration
The power rule for integration is akin to a trusty tool you keep in your math toolbox. It applies when you have a term of the form \(u^n\). For the integral \(\int u^n \, du\), the rule is: \[\int u^n \, du = \frac{u^{n+1}}{n+1} + C\]provided \(n eq -1\).
  • In our exercise, the term was \(u^{-1/2}\). Using the power rule, we integrated \(u^{-1/2}\) giving us \(2u^{1/2} + C\).
  • Remember, \(C\) is the constant of integration, appearing whenever you evaluate an indefinite integral.
  • This rule saves time and effort, especially when working with powers different than \(-1\).
Mastering this rule is crucial since it forms the basis for more complex integration techniques.
Definite and Indefinite Integrals
Integrals come in two forms: definite and indefinite. But what's the difference and when do you use each?
  • An indefinite integral, like the one in our example \(\int \frac{2x^2 \, dx}{\sqrt{2x^3 + 1}}\), represents a family of functions. It has an added constant \(C\), since integration is the reverse of differentiation and can't determine where exactly the function starts.
  • Conversely, a definite integral has limits, say from \(a\) to \(b\), and calculates the area under the curve between these points. There's no constant \(C\) here, as the result is a specific value.
When solving integrals, always note what you're after: a general form (indefinite) or a specific area/value (definite). Knowing this guides the entire integration process.

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