Integration of the Reciprocal Function

Expert reviewed • 05 March 2025 • 5 minute read

HSC Maths Advanced Syllabus

establish and use the formulae $\int\frac{1}{x}dx=ln|x|+C$ and $\int\frac{f'(x)}{f(x)}dx=ln|f(x)|+C$ for $x\ne0$ , $f(x)\ne0$ , respectively

Note:

Video coming soon!

The Reciprocal Function

As we know the reciprocal function is displayed by $\frac{d}{dx}log_ex=\frac{1}{x}$ . By reversing this we can determine the formula to integrate expressions in the form $\frac{1}{x}$ . Thus, the formula to integrate the reciprocal function, is as follows:

\int\frac{1}{x}dx=log_ex+C

However, the function $\frac{1}{x}$ can also be negative, and as such, has primitive $log_e(-x)$ as well as $log_ex$ . Thus, we manipulate the formula for integrating the reciprocal function to be:

\int\frac{1}{x}dx=log_e|x|+C

Additionally, there are two other standard formulas used for integrating the reciprocal function.

\int\frac{1}{ax+b}dx=\frac{1}a{}log_e|ax+b|+C

\int\frac{u'}{u}dx=log_e|u|+C \quad or \quad \int\frac{f'(x)}{f(x)}dx=log_e|f(x)|+C

Practice Question 1

Integrate the following function: $f(x)=\frac{x}{5x^2+1}$

Solution

To be able to solve the integral, we must first determine $f(x)$ and $f'(x)$ .

f(x)=5x^2+1\\f'(x)=2\times5x^{2-1}\\=10x

As we can see, the values we have determined, do not yet fit the formulas provided in the chapter above. Thus, we must manipulate the integral to fit the needed form.

\frac{1}{10}\int10\times\frac{x}{5x^2+1}dx

Now we can solve the integral:

\frac{1}{10}\int10\times\frac{x}{5x^2+1}dx=\frac{1}{10}\times log_e|5x^2+1|+C\\=\frac{1}{10}log_e|5x^2+1|+C

When we are asked to solve increasingly difficult integration questions, the process in which we we solve the question can vary slightly. In the video above, we explore this concept, looking at solving the integral: $\int_1^elog_exdx$

Integration of Exponential Functions

Calculus with Different Bases

Return to Module 5: Exponential and Logarithmic Functions