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Chapter 5: Problem 31

Evaluate the indefinite integral. \(\int x(2 x+5)^{8} d x\)

Short Answer

Expert verified
\(\frac{1}{40} (2x + 5)^{10} - \frac{5}{36} (2x + 5)^9 + C\)

Step by step solution

01

Recognize the Structure

Notice that the integrand is in the form of a product of a function and its derivative. Here, the function inside the integral is \((2x + 5)^8\), and its derivative is a constant multiple of \(x\). This suggests a substitution method for integration.
02

Choose a Substitution

Let \(u = 2x + 5\). Then, compute the derivative \(du/dx = 2\). Rearranging, we get \(du = 2 \, dx\) or \(dx = \frac{1}{2} du\). Also express \(x\) in terms of \(u\):\(x = \frac{u - 5}{2}\).
03

Substitute into the Integral

Replace all \(x\) and \(dx\) in the original integral:\[\int x(2x+5)^8 \, dx = \int \frac{u - 5}{2} u^8 \cdot \frac{1}{2} \, du\]Simplify this to:\[\frac{1}{4} \int (u - 5) u^8 \, du\]
04

Expand the Integrand

Distribute to expand the integrand:\[\frac{1}{4} \int (u^9 - 5u^8) \, du\]
05

Integrate Term by Term

Integrate each term separately:\[\frac{1}{4} \left( \int u^9 \, du - 5 \int u^8 \, du \right)\]This becomes:\[\frac{1}{4} \left( \frac{u^{10}}{10} - 5 \cdot \frac{u^9}{9} \right) + C\]
06

Simplify the Expression

Simplify the integrated result:\[\frac{1}{4} \left( \frac{u^{10}}{10} - \frac{5u^9}{9} \right)\]This can further be simplified to:\[\frac{1}{40} u^{10} - \frac{5}{36} u^9 + C\]
07

Substitute Back in Terms of x

Replace \(u\) with \(2x + 5\) to get the integral in terms of \(x\):\[\frac{1}{40} (2x + 5)^{10} - \frac{5}{36} (2x + 5)^9 + C\]
08

Final Result

The solution to the integral \(\int x(2x+5)^8 \, dx\) is:\[\frac{1}{40} (2x + 5)^{10} - \frac{5}{36} (2x + 5)^9 + C\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
In calculus, the substitution method is a popular technique for solving indefinite integrals, especially when you notice a function and its derivative in the integrand. The goal is to make the integration process easier by transforming a complicated integral into a simpler one.
To use substitution effectively:
  • Identify a part of the integrand (a function) to be replaced by a new variable, usually denoted as \( u \).
  • Derive \( du \) by differentiating \( u \) with respect to \( x \).
  • Solve for \( dx \) in terms of \( du \) or vice-versa. This allows you to replace the original variable with the substitute variable in the integral.
Once these substitutions are made, you perform the integral with respect to \( u \) instead of \( x \). After integrating, substitute back the expression for \( u \) in terms of \( x \) to get the answer in its original variable. This method is powerful because it can turn a complex integral into something much more manageable.
Integration Techniques
Integration techniques are strategies used to find the integral of a function. Besides substitution, several techniques aid in solving various types of integrals. Some of these techniques include:
  • Integration by Parts: This technique is useful when the integrand is a product of two functions, based on the product rule of differentiation.
  • Partial Fraction Decomposition: Used primarily for rational expressions, it involves expressing a complicated fraction as a sum of simpler fractions that are easier to integrate.
  • Trigonometric Integrals: For integrals involving trigonometric functions, identities and substitutions can simplify the process.
In our specific problem, the substitution method was applied because the structure of the integrand鈥攁n expression raised to a power coupled with its derivative鈥攎akes it an ideal candidate for substitution. Understanding and selecting the correct technique is crucial for efficiently solving any integral.
Integrand Expansion
Expanding the integrand is a common technique used to simplify the integration process. After making a substitution, you might end up with a product or composition of terms that are difficult to integrate. In such cases, expanding the expression into a sum of simpler terms is beneficial.
Here's how this helps:
  • The expansion process involves distributing across terms to transform a difficult expression into a series of simpler expressions that are straightforward to integrate.
  • Once expanded, you can integrate each term separately, which simplifies the overall computation.
In the given exercise, after substituting \( u = 2x + 5 \), we had an integrand \( \frac{1}{4} \int (u-5)u^8 \, du \). By expanding this to \( \frac{1}{4} \int (u^9 - 5u^8) \, du \), we broke down the task into integrating \( u^9 \) and \( 5u^8 \) separately. This step significantly simplifies the integration process.

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Most popular questions from this chapter

Computer algebra systems sometimes need a helping hand from human beings. Try to evaluate \(\int(1+\ln x) \sqrt{1+(x \ln x)^{2}} d x\) with a computer algebra system. If it doesn't return an answer, make a substitution that changes the integral into one that the CAS can evaluate.

\(69-78\) Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \(g(x)=\int_{1}^{x} \frac{1}{t^{3}+1} d t\)

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{2 \pi}^{\infty} \sin \theta d \theta\)

A manufacturer of lightbulbs wants to produce bulbs that last about 700 hours but, of course, some bulbs burn out faster than others. Let \(F(t)\) be the fraction of the company's bulbs that burn out before \(t\) hours, so \(F(t)\) always lies between 0 and \(1 .\) (a) Make a rough sketch of what you think the graph of \(F\) might look like. (b) What is the meaning of the derivative \(r(t)=F^{\prime}(t) ?\) (c) What is the value of \(\int_{0}^{\infty} r(t) d t ?\) Why?

Find the general indefinite integral. \(\int v\left(v^{2}+2\right)^{2} d v\)
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