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.

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}\ln x={\displaystyle \frac{1}{x}}$$

Using the properties of powers, we can rewrite the above equation.

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

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 $2x=u$.

$$2x=u\to \mathrm{d}x=\frac{1}{2}\mathrm{d}x$$

• 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 $\int \ln (\frac{x}{x+1})\mathrm{d}x$.

• Use the properties of logarithms to write the quotient as a difference.

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

• Use the properties of integrals to split the integral.

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

• Let $u=x+1$ so $\mathrm{d}x=\mathrm{d}u$ and substitute in the second integral.

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

• Integrate the natural logarithmic functions.

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

• Substitute back $u=x+1$.

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

• Use algebra to simplify.

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

• Simplify using the properties of logarithms.

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