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Chapter 7: Problem 34

Evaluate the integral. $$ \int \frac{\left(2^{x}+1\right)^{2}}{2^{x}} d x $$

Short Answer

Expert verified
\(\frac{2^x}{\ln(2)} + 2x - \frac{2^{-x}}{\ln(2)} + C\)

Step by step solution

01

Rewrite the Integrand

First, rewrite the integrand by expanding the numerator. The expression \((2^x + 1)^2\) expands to \(2^{2x} + 2 \cdot 2^x \cdot 1 + 1^2\), simplifying to \(2^{2x} + 2 \cdot 2^x + 1\). Thus, the integral becomes: \(int \left(\frac{2^{2x}}{2^x} + \frac{2 \cdot 2^x}{2^x} + \frac{1}{2^x}\right) dx\).
02

Simplify the Integrand

Simplify each term of the integrand: \(int \left(2^x + 2 + 2^{-x}\right) dx\). Notice that each term is now a simpler expression that can be integrated independently.
03

Integrate Each Term Separately

Integrate each term separately: - For \(2^x\), the integral is \(\frac{2^x}{\ln(2)}\).- For 2, the integral is \(2x\). - For \(2^{-x}\), the integral is \(-\frac{2^{-x}}{\ln(2)}\).Combine these results to find the integral of the original expression.
04

Combine Results and Add Constant of Integration

Combine the results from Step 3: \(\frac{2^x}{\ln(2)} + 2x - \frac{2^{-x}}{\ln(2)} + C\).Here, \(C\) is the constant of integration, which accounts for any constant that could have been differentiated to yield zero in the original integral.

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Key Concepts

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

Polynomial Expansion
Polynomial expansion involves breaking down expressions into a sum of terms based on existing mathematical rules and identities. In this integral problem, we start with \((2^x + 1)^2\).To expand this polynomial, we use the algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\).
For our specific polynomial:
  • Set \(a = 2^x\) and \(b = 1\).
  • Thus, \((2^x + 1)^2 = (2^x)^2 + 2(2^x)(1) + 1^2\).
  • This simplifies to \(2^{2x} + 2 imes 2^x + 1\).
Breaking down complex expressions using polynomial expansion makes integration and other calculations much more manageable. It allows us to convert a complicated expression into more straightforward terms that are easier to work with individually.
Integration Techniques
Integration techniques are essential tools in calculus for finding the antiderivative or integral of a function. In our example, the expression\(\frac{(2^x + 1)^2}{2^x}\)is simplified into three separate terms: \(2^x, 2,\) and \(2^{-x}\).
To integrate the function:
  • The integral of \(2^x\) is \(\frac{2^x}{\ln(2)}\)because \(\ln(2)\) is the constant of integration for exponential functions with base 2.
  • The integral of the constant \(2\) is straightforward as \(2x\).
  • The integral of \(2^{-x}\) requires the understanding of negative exponents, resulting in \(-\frac{2^{-x}}{\ln(2)}\).
These integration techniques make working with both simple and complex functions in calculus more feasible by breaking them down into manageable parts.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions and are crucial in integration, especially with non-standard bases like 2. In our integral example, logarithmic functions play a key role in calculating the antiderivative for \(2^x\) and \(2^{-x}\).
For base 2 exponential functions, the natural logarithm \(\ln(2)\) is used in the denominator of the antiderivative. This is because:
  • The derivative of \(2^x\) is proportional to itself but multiplied by \(\ln(2)\),which means the antiderivative will include a division by \(\ln(2)\).
  • Similarly, \(2^{-x}\) is integrated by including \(-\frac{1}{\ln(2)}\) since the negative exponent switches the direction of the growth.
Understanding how logarithmic functions interact with exponents enriches the understanding and application of integral calculus, making it easier to solve even complex integrals.

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