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Magazine Chapter 7: Problem 101
Evaluate the integrals. $$\int_{0}^{2} \frac{\log _{2}(x+2)}{x+2} d x$$
Short Answer
Expert verified
The integral evaluates to \( \frac{3\ln(2)}{2} \).
Step by step solution
01
Change of Variable
To simplify the integral, let's use a change of variable: let \( u = x + 2 \). Then \( du = dx \). When \( x = 0 \), \( u = 2 \), and when \( x = 2 \), \( u = 4 \). The integral becomes \[ \int_{2}^{4} \frac{\log_{2}(u)}{u} \, du \].
02
Convert Logarithm Base
The natural logarithms are often easier to work with. Convert \( \log_2(u) \) to natural logarithm using the change of base formula: \( \log_2(u) = \frac{\ln(u)}{\ln(2)} \). Substitute this back into the integral to get \[ \int_{2}^{4} \frac{\ln(u)/\ln(2)}{u} \, du = \frac{1}{\ln(2)} \int_{2}^{4} \frac{\ln(u)}{u} \, du \].
03
Integrate by Parts
To evaluate the integral \( \int \frac{\ln(u)}{u} \, du \), consider integrating by parts. Let \( v = \ln(u) \) and \( dw = \frac{1}{u} \, du \). Then \( dv = \frac{1}{u} \, du \) and \( w = \ln(u) \). Thus, the integrals by parts formula \( \int v \, dw = vw - \int w \, dv \) gives us: \[ \int \frac{\ln(u)}{u} \, du = \ln(u) \cdot \ln(u) - \int \ln(u) \cdot \frac{1}{u} \cdot du = \frac{1}{2} \ln^2(u) \].
04
Evaluate the Definite Integral
With \( \frac{1}{(2)} \int_{2}^{4}{\frac{\ln(u)}{u} \, du} = \frac{1}{2\ln(2)} [\ln^2(u)]_{2}^{4} \) evaluated as: \[\frac{1}{2\ln(2)} ( (\ln(4))^2 - (\ln(2))^2 ) = \frac{1}{2\ln(2)} ( (\ln(2^2))^2 - (\ln(2))^2 ) = \frac{1}{2\ln(2)} ( (2\ln{2})^2 - (\ln{2})^2 ) = \frac{1}{2\ln(2)} ( 4\ln^2(2) - 1\ln^2(2) ) = \frac{3\ln(2)}{2} \].
05
Final Solution
The value of the integral \( \int_{0}^{2} \frac{\log _{2}(x+2)}{x+2} \, d x \) is \( \frac{3\ln(2)}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Change of Variable
Change of variable, often referred to as substitution, is a powerful method in integral calculus for simplifying an integral by transforming it into a new variable, typically making the integration more straightforward. In our exercise, we're given the integral \( \int_{0}^{2} \frac{\log _{2}(x+2)}{x+2} \, dx \). By choosing \( u = x + 2 \), we change the variable to make the expression more manageable.
- This substitution shifts the limits of integration from \( x = 0 \) to \( x = 2 \) into \( u = 2 \) to \( u = 4 \).
- Here, \( du = dx \), simplifying our computations as there is no extra factor introduced by this transformation.
Integration by Parts
Integration by parts is a technique derived from the product rule for differentiation which is often used when integrating products of two functions that are difficult to integrate directly. The formula is given by:
- \( \int v \, dw = vw - \int w \, dv \)
- \( dv = \frac{1}{u} \, du \)
- \( w = \ln(u) \)
Logarithmic Integration
Logarithmic integration is crucial here as the integrand includes a logarithmic function, specifically \( \log_{2} \) in the original problem. Employing the change of base formula:
- \( \log_2(u) = \frac{\ln(u)}{\ln(2)} \)
