Primitives of continuous functions play a key role in the calculation of integrals. Therefore, in this article, we will use examples to introduce a simple technique for determining continuous function primitives.

We assume that the reader knows how to calculate integrals.

## Generalities on primitives of continuous functions

Let $f:D\subset \mathbb{R}$ be a continuous function. The primitive of the function $f$ is a differentiable function $F:D\to\mathbb{R}$ such that $F'(x)=f(x)$ for any $x\in D$.

If $f$ admits a primitive $F$, then the function $G=F+c$ is a primitive of $f$ for any constant $c\in\mathbb{R}$. Two primitives of $f$ differ from a constant.

Let us now state the fundamental theorem of calculus: Let $f:D\to \mathbb{R}$ be a continuous function. The n for any $a\in D,$ the function defined by We also write \begin{align*}F(x)=\int^x_a f(t)dt,\quad x\in D,\end{align*} is differentiable on $D$ and $Fâ=f$.

Remark that the primitive a function of $C^1$ class. This means that the derivative function $Fâ$ is continuous.

## Examples of how to find primitivesÂ

To determine a primitive of a continuous function, you need to know some classical primitives as well as partial fraction decomposition rules.

- Determine a primitive of the function \begin{align*} f(x)=\frac{2x^2-3x+4}{(x-1)^2}\textrm{ on }]1,+\infty[ .\end{align*} We look for real constants $a,b$ and $c$ such that \begin{align*} f(x)=a+\frac{b}{x-1}+\frac {c}{(x-1)^2}. \end{align*} By putting everything at the same denominator in the right-hand side, we find \begin{align*} f(x)=\frac{ax^2+x(-2a+b)+(a-b+c)}{(x-1)^2}. \end{align*} According to the first expression of $f,$ one obtains $a=2,;b=1$ and $c=3$, so that for any $x>1,$ \begin{align*} f(x)=2+\frac{1}{x-1}+\frac {3}{(x-1)^2}. \end{align*} The primitive of $f$ on $]1,+\infty[$ is then given by \begin{align*} F(x)=2x+\ln(x-1)-\frac {3}{x-1}+C,\qquad C\in \mathbb{R} \end{align*}
- Determine the primitive of the function \begin{align*} f(x)=\frac{2x-1}{1+x^2}\textrm{ on }\mathbb{R}.\end{align*} We first observe that $f$ is defined on $\mathbb{R}$. On the other hand, we can write \begin{align*} f(x)&=\frac{2x}{1+x^2}-\frac{1}{1+x^2}\cr &=\frac{(1+x^2)â}{1+x^2}-\frac{1}{1+x^2}\cr &= \frac{d}{dx}\left(\ln(1+x^2)-\arctan(x)\right). \end{align*} Thus, the primitive of $f$ on $\mathbb{R}$ is given by \begin{align*} F(x)=\ln(1+x^2)-\arctan(x)+C,\qquad C\in\mathbb{R}. \end{align*}
- Calculate the primitive of the following function\begin{align*} f(x)=x e^{\lambda x}\textrm{ on }\mathbb{R}.\end{align*} In this question, we will use integration by parts method. We then have for any $\lambda\neq 0,$ \begin{align*} \int^x_c t e^{\lambda t}dt&= \int^x_c \frac{t}{\lambda} \left(e^{\lambda t}\right)âdt\cr & =\left[\frac{t}{\lambda} e^{\lambda t}\right]^x_c-\int^x_c \frac{1}{\lambda}e^{\lambda t}dt. \end{align*} Hence the primitive of $f$ on $\mathbb{R}$ is given by \begin{align*} F(x)=\frac{1}{\lambda} xe^{\lambda x}-\frac{1}{\lambda^2} e^{\lambda x}+C,\qquad C\in\mathbb{R}. \end{align*}
- Compute the primitive of the function\begin{align*} f(x)=\frac{1}{\sqrt{x^2+2x+5}}\textrm{ on }\mathbb{R}.\end{align*} Observe that $f$ is defined on all $\mathbb{R}$. Here we shall use the change of variables technique. We have \begin{align*} \int^x_c \frac{dt}{\sqrt{t^2+2t+5}}&=\int^x_c \frac{dt}{\sqrt{(t^2+2t+4)+1}}\cr &=\int^x_c \frac{dt}{\sqrt{(t+2)^2+1}}. \end{align*} We now put $u=t+2$ then $du=dt$ and \begin{align*} \int^x_c \frac{dt}{\sqrt{t^2+2t+5}}&=\int^{x+2}_{c+2} \frac{du}{\sqrt{u^2+1}}\cr &= 2\left[\ln(u+\sqrt{u^2+1})\right]^{x+2}_{c+2}\cr &= 2\ln(x+2+\sqrt{x^2+2x+5})+C. \end{align*} This implies that the primitive of $f$ is \begin{align*} F(x)=2\ln(x+2+\sqrt{x^2+2x+5})+C. \end{align*}