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Chapter 7: Problem 55

Logarithm base \(b\) Prove that $$ \int \log _{b} x d x=\frac{1}{\ln b}(x \ln x-x)+C $$

Short Answer

Expert verified
Question: Prove that $$\int \log _{b} x dx = \frac{1}{\ln b}(x \ln x-x)+C$$. Answer: Using the change of base formula, integration by parts, and simplification, we have proven that: $$\int \log _{b} x dx = \frac{1}{\ln b}(x \ln x-x)+C$$.

Step by step solution

01

Apply the change of base formula

First, remember that we can apply the change of base formula to rewrite the logarithm as follows: $$\log_{b}x = \frac{\log x}{\log b}$$ Now the integral becomes: $$\int \frac{\log x}{\log b} dx$$ Notice that the \(\frac{1}{\log b}\) does not involve \(x\) so it can be treated as a constant factor and brought out of the integral. Thus, this integral can now be reformulated as: $$\frac{1}{\log b} \int \log x dx$$
02

Use integration by parts

We will now use integration by parts to solve the remaining integral. Let's pick the following functions: $$ u = \log x \quad \Rightarrow \quad du = \frac{1}{x} dx\\ dv = dx \quad \Rightarrow \quad v = x $$ Now we can apply the integration by parts formula: $$\frac{1}{\log b} \int \log x dx = \frac{1}{\log b}(u v - \int v du)$$ Plugging in the functions we chose and their derivatives/integrals, we get: $$\frac{1}{\log b} \int \log x dx = \frac{1}{\log b}(x \log x - \int x \cdot \frac{1}{x} dx)$$
03

Simplify the integral and integrate

Now we can simplify the integral and solve: $$\frac{1}{\log b}(x \log x - \int x \cdot \frac{1}{x} dx) = \frac{1}{\log b}(x \log x - \int dx)$$ Integrating the remaining integral: $$\frac{1}{\log b}(x \log x - \int dx) = \frac{1}{\log b}(x \log x - x)$$
04

Add the constant of integration

Finally, we have to include the constant of integration, \(C\). Our final answer is: $$ \frac{1}{\log b}(x \log x - x) + C = \frac{1}{\ln b}(x \ln x-x)+C $$ Hence, we have proven that: $$\int \log _{b} x dx = \frac{1}{\ln b}(x \ln x-x)+C$$

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Key Concepts

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

Change of Base Formula
When working with logarithms, converting to a familiar base can make calculations easier. This is especially true in integration problems. The change of base formula allows us to express any logarithm \(\log_{b}x\) in terms of natural logarithms, making it compatible with most calculus operations. The change of base formula is given by:
  • \(\log_{b}x = \frac{\log x}{\log b}\)
This formula is very useful because it transforms logarithms into the form where the natural logarithm, \(\log x\), is the primary function.
By applying this formula, we can simplify logarithmic expressions into a form that is readily integrable, which is crucial in solving complex integrals.
This step was the first in transforming the original integral problem to apply more straightforward and standard integration techniques like integration by parts.
Logarithm Integration
Integrating a logarithmic function can seem tricky at first, but using the right techniques makes the process more manageable. Here, integration by parts, a method of substituting parts of the integrand to make integration possible, is key. The integration by parts formula is:\[\int u \, dv = uv - \int v \, du\]To integrate \(\log x\), choose:
  • \(u = \log x\)
  • \(dv = dx\)
This choice implies that \(du = \frac{1}{x}dx\) and \(v = x\). Using these substitutions, the integral becomes:\[x \, \log x - \int x \, \frac{1}{x} \, dx\]This simplifies to \(x \, \log x - \int dx\), which is straightforward to integrate since it only involves basic integration of \(x\).
The result from the integration will include the initial logarithmic term adjusted with constants derived from the original logarithm's base.
Constant of Integration
In calculus, particularly integral calculus, the constant of integration, often denoted as \(C\), is essential to express the most general form of an antiderivative. When solving indefinite integrals, the constant appears because differentiation of a constant results in zero.
Therefore, an infinite number of functions could satisfy the integral, and to capture all these potential solutions, we add the constant of integration.In our result:\[\frac{1}{\ln b}(x \ln x - x) + C\]The \(C\) is added to indicate the family of possible functions that could be integrated to produce the specified derivative.
Neglecting to include the constant in solutions results in incomplete representations of the solution set. Thus, always remember the constant of integration when answering indefinite integrals! This ensures that every potential antiderivative is accounted for, reinforcing the comprehensive nature of calculus solutions.

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Most popular questions from this chapter

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-\alpha x^{2}}\). a. Graph the Gaussian for \(a=0.5,1,\) and 2 b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.

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