Integration of Logarithmic Functions

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!

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.

$$\dfrac{\mathrm{d}}{\mathrm{d}x}\ln{x}=x^{\text{-}1}$$

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!

$$\int(x^4-2x^2+x^{\text{-}1}-x^{\text{-}3})\mathrm{d}x=\frac{1}{5}x^5-\frac{2}{3}x^3+\ln{|x|}+\frac{1}{2}x^{\text{-}2}+C$$

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.

$$\int\log_{2}{x}\,\mathrm{d}x=\frac{1}{\ln{2}}\int\ln{x}\,\mathrm{d}x$$

• Integrate the natural logarithmic function.

$$\int\log_{2}{x}\,\mathrm{d}x=\frac{1}{\ln{2}}x(\ln{x}-1)+C$$

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.

$$\int \ln{2x} \,\mathrm{d}x = \frac{1}{2} \int \ln{u} \mathrm{d}u$$

• Integrate the natural logarithmic function.

$$\int \ln{2x} \,\mathrm{d}x = \frac{1}{2} u(\ln{u}-1)+C$$

• Substitute back \( 2x=u \) into the integral and simplify.

$$\int \ln{2x} \,\mathrm{d}x= \frac{1}{2} (2x)(\ln{2x}-1)+C$$

$$\int \ln{2x} \,\mathrm{d}x = x(\ln{2x}-1)+C$$

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.

$$ \int \ln{x^2} \mathrm{d}x = 2 \int \ln{x} \mathrm{d}x$$

• Integrate the natural logarithmic function.

$$ \int \ln{x^2} \mathrm{d}x = 2x(\ln{x}-1) + C$$

We can also distribute 2 inside and use it to write it as an exponent in the natural logarithm.

$$ \int \ln{x^2} \mathrm{d}x = x( 2\ln{x} -2) + C$$

$$ \int \ln{x^2} \mathrm{d}x = x(\ln{x^2} -2) + C$$

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 \( u=x+1 \).

$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x(\ln{x}-1)+(x+1)(\ln{(x+1)}-1)+C$$

• Use algebra to simplify.

$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x\ln{x} - x\ln{(x+1)} - \ln{(x+1)}$$

• Simplify using the properties of logarithms.

$$ \int \ln{\Big(\frac{x}{x+1}\Big)} \mathrm{d}x = x\ln{\Big(\frac{x}{x+1}\Big)}-\ln{(x+1)} $$