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Chapter 6: Problem 14

Find the indefinite integral. $$ \int \frac{x}{\sqrt[5]{2 x^{2}+5}} d x $$

Short Answer

Expert verified
\( \frac{5}{16} (2x^2 + 5)^{4/5} + C \)

Step by step solution

01

Identify the Substitution

To solve the integral \( \int \frac{x}{\sqrt[5]{2x^2+5}}\, dx \), we need to make a substitution that simplifies the integrand. Notice that \( 2x^2 + 5 \) is under a root, so let \( u = 2x^2 + 5 \). This substitution will help manage the complexity of the root.
02

Differentiate the Substitution

Differentiate the substitution \( u = 2x^2 + 5 \) with respect to \( x \) to find \( \frac{du}{dx} \). This gives \( du = 4x\, dx \). We can solve for \( x\, dx \) as \( x\, dx = \frac{1}{4}du \).
03

Rewrite the Integral

Substitute \( u = 2x^2 + 5 \) and \( x\, dx = \frac{1}{4}du \) into the integral: \[ \int \frac{x}{\sqrt[5]{2x^2 + 5}}\, dx = \int \frac{1}{4} \cdot \frac{1}{u^{1/5}}\, du. \] This simplifies to \[ \frac{1}{4} \int u^{-1/5}\, du. \]
04

Integrate

Perform the integration: \( \frac{1}{4} \int u^{-1/5} \, du \). Use the power rule for integration \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Here, \( n = -\frac{1}{5} \). The integral becomes \[ \frac{1}{4} \cdot \frac{u^{4/5}}{4/5} + C = \frac{1}{4} \cdot \frac{5}{4} u^{4/5} + C \] \[ = \frac{5}{16} u^{4/5} + C. \]
05

Substitute Back

Substitute back \( u = 2x^2 + 5 \) into the integrated expression: \[ \frac{5}{16} (2x^2 + 5)^{4/5} + C. \] This gives the final indefinite integral solution.

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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 technique used to simplify complex integrals, making them easier to evaluate. It involves substituting a part of the integrand with a new variable鈥攃ommonly denoted as \( u \). This transformation usually reduces the complexity of the integral, turning it into a more familiar form.To use the substitution method, follow these steps:
  • Identify a portion of the integrand that can be replaced by a simpler expression. Look for expressions that appear in the derivative to simplify calculations.
  • Make a substitution, such as \( u = g(x) \), where \( g(x) \) is the chosen portion of the integrand.
  • Differentiate the substitution to express \( dx \) in terms of \( du \). This can be written as \( du = g'(x)\, dx \), and solve for \( dx \).
  • Rewrite the original integral in terms of \( u \) and \( du \) and perform the integration.
  • Finally, substitute back the original expression for \( u \) to get the solution in terms of the original variable.
In the example exercise, the substitution \( u = 2x^2 + 5 \) simplifies the integral by removing the root complexity, allowing for a straightforward integration.
Power Rule for Integration
The power rule for integration is a simple yet powerful technique to find integrals of expressions with powers. When applying the power rule, the integral of \( u^n \) with respect to \( u \) is calculated using the formula:\[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \]where \( n eq -1 \) and \( C \) is the constant of integration. Here's how it works:
  • Add 1 to the exponent \( n \) of the variable \( u \).
  • Divide by the new exponent \( n + 1 \).
  • Don't forget to add the constant of integration \( C \).
In our exercise, the integrand transformed to \( u^{-1/5} \). Applying the power rule, we integrate to get:\[ \frac{u^{4/5}}{4/5} + C \]Simplifying gives us the result \( \frac{5}{4} u^{4/5} + C \). This showcases how the power rule simplifies integration of expressions with variable powers.
Integration Techniques
Integration techniques are vital for tackling a wide range of functions you may encounter in calculus. Various techniques exist, each suitable for different kinds of problems. Here, we encountered two commonly used techniques: substitution and the power rule. Let's explore them further.
  • Substitution Method: Often used for integrals involving composites or nested functions. It simplifies the integral by changing the variable, making awkward expressions easier to handle.
  • Power Rule for Integration: A straightforward technique for polynomials. If the integrand is a simple polynomial in standard form, apply the power rule as a first choice.
There are other techniques too:
  • Partial Fraction Decomposition: Useful for rational functions, splitting complex fractions into simpler ones before integrating.
  • Integration by Parts: Efficient for products of functions. It uses the formula \( \int u \, dv = uv - \int v \, du \) to integrate pieces separately.
  • Trigonometric Substitution: Applies primarily to integrals involving square roots and trigonometric identities.
Understanding when and how to apply these techniques is crucial. In this exercise, the combination of substitution and power rule effectively simplified and solved the integral.

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