Have you ever wondered how noise is measured? You might have noticed that your perception of noise depends on where you are standing. It is not the same noise you hear in a busy street when walking down the sidewalk as being inside a car.
Traffic in a busy street - pixabay.com
To measure noise, scientists use a logarithmic scale. This scale takes into account different factors, like position and the source of the noise. Many calculations related to the measure of sound and noise involve logarithms in a certain way, and integrals are not the exception. Here we will learn how to deal with integrals that involve logarithmic functions.
Integrals Involving Logarithmic Functions
One of the most essential differentiation rules is The Power Rule, which lets us differentiate any power function. Because of this rule, the integration of a power function is as straightforward as its derivative. Let's recall this with a quick example.
Evaluate the integral \(\int x^2\mathrm{d}x\).
To integrate a power function we need to increase the power of the variable by 1 and divide by the value of the new power.
$$\int x^2\mathrm{d}x=\frac{1}{2+1}x^{2+1}+C$$
$$\int x^2\mathrm{d}x=\frac{1}{3}x^3+C$$
Pretty simple, right? This also works with negative powers!
$$\int x^{\text{-}2}\mathrm{d}x=\frac{1}{-2+1}x^{-2+1}+C$$
$$\int x^{\text{-}2}\mathrm{d}x=\frac{1}{-1}x^{-1}+C$$
$$\int x^{\text{-}2}\mathrm{d}x=\text{-}\frac{1}{x}+C$$
But what if we try to apply this rule when the power is equal to \(\text{-}1\)?
$$\int x^{\text{-}1}\mathrm{d}x=\frac{1}{-1+1}x^{-1+1}+C$$
We would be dividing by zero, which is not allowed!
The integration of power rule does not apply when the power is \(\text{-}1\). Luckily, there is a way around it.
We will now recall how to differentiate a natural logarithm function.
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln{x}=\dfrac{1}{x}$$
Using the properties of powers, we can rewrite the above equation.
$$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln{x}=x^{\text{-}1}$$
In the above example, we found how to deal with a power equal to \(\text{-}1\). We are now ready to write the antiderivative of the power function when the power is equal to \(\text{-}1\).
The antiderivative of a power function when its power is equal to \(\text{-}1\) is the absolute value of the natural logarithmic function.
$$\int x^{\text{-}1}\mathrm{d}x=\ln{|x|}+C$$
You may also find the above formula written in one of the following ways:
$$\int\frac{1}{x}\mathrm{d}x=\ln{|x|}+C$$
$$\int\frac{\mathrm{d}x}{x}=\ln{|x|}+C$$
Wait, why is there an absolute value? It turns out that the domain of the natural logarithmic function consists of positive numbers only. Since the function \( \frac{1}{x} \) can take negative values we need to include both scenarios, which can be summarized by taking the absolute value.
Thanks to this formula, integral of all power functions becomes pretty straightforward.
Evaluate the following integral:
$$\int(x^4-2x^2+x^{\text{-}1}-x^{\text{-}3})\mathrm{d}x$$
We begin by noting that the integral can be split into 4 integrals.
$$\int(x^4-2x^2+x^{\text{-}1}-x^{\text{-}3})\mathrm{d}x=\int x^4\mathrm{d}x-2\int x^2\mathrm{d}x+\int x^{\text{-}1}\mathrm{d}x-\int x^{\text{-}3}\mathrm{d}x$$
We can now integrate each power function individually. Don't forget to add the integration constant!
$$\int(x^4-2x^2+x^{\text{-}1}-x^{\text{-}3})\mathrm{d}x=\frac{1}{5}x^5-\frac{2}{3}x^3+\ln{|x|}+\frac{1}{2}x^{\text{-}2}+C$$
Integration of Logarithmic Functions Formula
And what if we need to integrate the natural logarithmic function? It turns out that the antiderivative of the natural logarithm function is not straightforward, but still, let's take a look at it.
Ready to take a dive into the proof? Let's go!
As stated before, the natural logarithmic function is not the derivative of any simple function, so integrating it is not straightforward.
$$\int\ln{x}\,\mathrm{d}x$$
In these cases we have to resort to different Integration Techniques. Let's try using Integration by Parts.
$$u=\ln{x} \rightarrow \mathrm{d}u=\frac{\mathrm{d}x}{x}$$
$$\mathrm{d}v=\mathrm{d}x \rightarrow v=x$$
We now use the Integration by Parts formula \(\int u\mathrm{d}v=uv-\int v\mathrm{d}u\).
$$\int\ln{x}\,\mathrm{d}x=(\ln{x})(x)-\int x\cdot\frac{\mathrm{d}x}{x}$$
Rewrite the above expression, so it looks clearer and nicer.
$$\int\ln{x}\,\mathrm{d}x=x\ln{x}-\int\mathrm{d}x$$
Remember that the integral of \(\mathrm{d}x\) is just \(x\). Do not forget to add the integration constant!
$$\int\ln{x}\,\mathrm{d}x=x\ln{x}-x+C$$
Finally, we factor \(x\) and we are done!
$$\int\ln{x}\,\mathrm{d}x=x(\ln{x}-1)+C$$
What if the logarithm has a base other than \(e\)? We can use the properties of logarithms!
A logarithm with a base other than \(e\) can be rewritten using the following property:
$$\log_{a}{x}=\frac{\ln{x}}{\ln{a}}$$
Where \(a\) is a real number greater than \(0\).
Our strategy for dealing with logarithms is to rewrite them in terms of natural logarithms. Let's see how to use this property on an integral.
Evaluate the integral \( \int\log_{2}{x}\,\mathrm{d}x \).
- Use the properties of logarithms to rewrite the integrand.
$$\int\log_{2}{x}\,\mathrm{d}x=\int\frac{\ln{x}}{\ln{2}}\,\mathrm{d}x$$
Note that \( \ln{2} \) is a constant so that it can be taken out of the integral.
- Take out \( \frac{1}{\ln{2}} \) out of the integral.
$$\int\log_{2}{x}\,\mathrm{d}x=\frac{1}{\ln{2}}\int\ln{x}\,\mathrm{d}x$$
$$\int\log_{2}{x}\,\mathrm{d}x=\frac{1}{\ln{2}}x(\ln{x}-1)+C$$
After looking at the above example, we can generalize the integration for logarithmic functions formula to include any kind of logarithms.
The antiderivative of a logarithmic function is given by:
$$\int\log_{a}{x}\, \mathrm{d}x=\frac{1}{\ln{a}}x(\ln{x}-1)+C$$
Where \(a\) is a real number greater than \(0\).
Integration of Logarithmic Functions Examples
The best way to get better at integration is by practicing! Let's see more examples of integrals involving logarithmic functions.
Evaluate the integral \( \int \ln{2x}\, \mathrm{d}x \).
We can evaluate this integral easily by doing the substitution \(2x=u\).
$$2x=u \rightarrow \mathrm{d}x=\frac{1}{2}\mathrm{d}x$$
- Substitute \(2x=u\) and \( \mathrm{d}x=\frac{1}{2}\mathrm{d}u\) into the integral.
$$\int\ln{2x}\,\mathrm{d}x=\int\ln{u} \cdot \frac{1}{2}\mathrm{d}u$$
- Take \(\frac{1}{2}\) out of the integral.
$$\int \ln{2x} \,\mathrm{d}x = \frac{1}{2} \int \ln{u} \mathrm{d}u$$
$$\int \ln{2x} \,\mathrm{d}x = \frac{1}{2} u(\ln{u}-1)+C$$
- Substitute back \( 2x=u \) into the integral and simplify.
$$\int \ln{2x} \,\mathrm{d}x= \frac{1}{2} (2x)(\ln{2x}-1)+C$$
$$\int \ln{2x} \,\mathrm{d}x = x(\ln{2x}-1)+C$$
Be sure to use any relevant properties of logarithms before integrating.
Evaluate the integral \( \int \ln{x^2} \mathrm{d}x \).
At first glance, we might think this is a complicated integral because it involves \( x^2 \), but there is no \( 2x \) to use integration by substitution. However, by using the properties of logarithms, this becomes fairly easy.
- Use the property of logarithms \( \ln{a^b} = b \ln{a} \) to rewrite the integral.
$$ \int \ln{x^2} \mathrm{d}x = \int 2 \ln{x} \mathrm{d}x$$
- Take \( 2 \) out of the integral.
$$ \int \ln{x^2} \mathrm{d}x = 2 \int \ln{x} \mathrm{d}x$$
$$ \int \ln{x^2} \mathrm{d}x = 2x(\ln{x}-1) + C$$
We can also distribute 2 inside and use it to write it as an exponent in the natural logarithm.
$$ \int \ln{x^2} \mathrm{d}x = x( 2\ln{x} -2) + C$$
$$ \int \ln{x^2} \mathrm{d}x = x(\ln{x^2} -2) + C$$
Here we will use the quotient property of logarithms.
Evaluate \( \int \ln{\Big(\frac{x}{x+1}\Big)} \, \mathrm{d}x \).
- Use the properties of logarithms to write the quotient as a difference.
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = \int (\ln{x}-\ln{(x+1)}) \, \mathrm{d}x $$
- Use the properties of integrals to split the integral.
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = \int \ln{x} \, \mathrm{d}x - \int \ln{(x+1)} \, \mathrm{d}x $$
- Let \( u=x+1\) so \(\mathrm{d}x=\mathrm{d}u \) and substitute in the second integral.
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = \int \ln{x} \, \mathrm{d}x - \int \ln{u} \, \mathrm{d}u $$
- Integrate the natural logarithmic functions.
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x(\ln{x}-1) + u(\ln{u}-1)+C$$
- Substitute back \( u=x+1 \).
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x(\ln{x}-1)+(x+1)(\ln{(x+1)}-1)+C$$
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x\ln{x} - x\ln{(x+1)} - \ln{(x+1)}$$
- Simplify using the properties of logarithms.
$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x\ln{\Big(\frac{x}{x+1}\Big)}-\ln{(x+1)} $$
Integrals Involving Logarithmic Functions - Key takeaways
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