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Find the following indefinite and definite integrals. \(\int x\left(x^{2}+1\right)^{100} d x\)

Short Answer

Expert verified
The indefinite integral is \( \frac{1}{202} (x^2 + 1)^{101} + C \).

Step by step solution

01

Identify the Integration Technique

The function inside the integral is the product of a polynomial and another polynomial raised to a power. We can use substitution to simplify the integration.
02

Choose a Substitution

Letting the inner function be the substitute, set \( u = x^2 + 1 \). Then, the derivative \( du = 2x \, dx \) can be rearranged as \( x \, dx = \frac{1}{2} du \), which will replace the \( x \, dx \) in the integral.
03

Rewrite the Integral

Substitute \( x^2 + 1 = u \) and \( x \, dx = \frac{1}{2} du \) into the integral. The expression becomes: \( \int x(x^2+1)^{100} \, dx = \int (u)^{100} \left(\frac{1}{2} du\right) = \frac{1}{2} \int u^{100} \, du \).
04

Integrate with Respect to u

Use the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Apply this rule to get: \( \frac{1}{2} \cdot \frac{u^{101}}{101} = \frac{1}{202} u^{101} + C \).
05

Substitute Back to Original Variables

Replace \( u \) with \( x^2 + 1 \) from the original substitution to get back in terms of \( x \): \( \frac{1}{202} (x^2 + 1)^{101} + C \).

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Key Concepts

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

Definite Integrals
In calculus, definite integrals are used to find the exact area under a curve over a specific interval. Unlike indefinite integrals, they result in a numerical value.
For a function \( f(x) \), integrating from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). This operation calculates the net area between the horizontal x-axis and the curve \( f(x) \), from x = a to x = b.
When dealing with definite integrals, one often uses the Fundamental Theorem of Calculus, which connects integration with differentiation. It's important to note:
  • Definite integrals can have positive, negative, or zero outcomes.
  • The limits \((a, b)\) make a definite integral different from an indefinite one.
  • To solve, you compute the antiderivative first, then evaluate it at the upper and lower limits and subtract.
Hence, definite integrals are extremely useful in many real-life applications, such as finding areas and in physics for calculating quantities like work and energy.
Integration by Substitution
Integration by substitution is a powerful method used to simplify integrals, especially when they involve complex functions. It often turns a difficult integral into a simpler one. This method is similar to the reverse of the chain rule in differentiation.
Here’s how it generally works:
  • Identify the part of the integral that is a composite function, usually noted as \( g(x) \).
  • Substitute \( u = g(x) \) to transform the integral. The goal is to replace complicated expressions with simpler ones.
  • Find \( du \), which represents the derivative of \( u \) with respect to \( x \), helping replace \( dx \) in the integral.
  • Rewrite the entire integral in terms of \( u \), simplifying the original expression.
This method makes it easier to integrate complex functions, as seen in problems like \( \int x(x^2+1)^{100} \ dx \), where letting \( u = x^2 + 1 \) guides us to a simplified solution.
Power Rule for Integration
The power rule for integration is analogous to the power rule for differentiation. It provides a straightforward way to integrate functions of the form \( x^n \), where \( n eq -1 \). The rule is:

\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]
Here, \( C \) is the constant of integration. This rule is fundamental because:
  • It simplifies the process of finding antiderivatives for polynomials.
  • It's applicable not only to simple powers of \( x \) but also to more complex expressions through substitution.
  • When combined with substitution, the power rule can tackle integrals involving polynomial expressions raised to a power, as in the example \( \int u^{100} \, du \).
Ultimately, mastering the power rule and its application through substitution is crucial for successfully solving a wide range of integration problems.

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