Antiderivatives of some known functions

Integration: Antiderivatives

Antiderivatives of some known functions

We have seen the antiderivatives of power functions. We will now discuss the antiderivatives of some other known functions.

$\int \frac{1}{x} \;\dd x= \ln |x| + \green C$

Example

$\begin{array}{rcl} \displaystyle \int \frac{6}{x} \;\dd x&=&6\cdot \displaystyle \int \frac{1}{x} \;\dd x \\ &=& 6 \cdot \ln |x| + \green C \end{array}$

Absolute value

Here $|x|$ is the absolute value of $x$ . This means

$|x|=\begin{cases}x&\text{if } x \geq 0\\ -x&\text{if } x \lt 0 \end{cases}$

$\int \e^x \;\dd x=\e^x + \green C$

Example

$\begin{array}{rcl} \displaystyle \int 3 \cdot \e^ x \;\dd x&=& 3 \displaystyle \int \e^x \; \dd x \\&=& 3 \cdot \e^x + \green C \end{array}$

$\int \blue a^x \;\dd x= \frac{\blue a^x}{\ln(\blue a)} + \green C$

Example

$\begin{array}{rcl} \displaystyle \int \blue {3}^ x \;\dd x&=& \dfrac{\blue 3^x}{\ln(\blue 3)} + \green C \end{array}$

$\int \ln(x) \; \dd x= {x \ln (x) -x }+ \green C$

Example

$\begin{array}{rcl} \displaystyle \int 4 \ln(x)\;\dd x &=& 4 \displaystyle \int \ln(x) \; \dd x \\ &=& 4 x \ln (x) - 4 x + \green C \end{array}$

$\int \log_\blue {a}(x) \; \dd x= \frac{x \ln (x) -x }{\ln (\blue a) }+ \green C$

Example

$\begin{array}{rcl} \displaystyle \int \log_\blue {7}(x)\;\dd x &=& \dfrac{x \ln (x) -x }{\ln (\blue 7) }+ \green C \end{array}$

$\int \sin(x) \; \dd x = -\cos(x) + \green C$

Example

$\begin{array}{rcl} \displaystyle\int 5 \cdot \sin(x) \; \dd x &=&\displaystyle5 \cdot \int \sin(x) \; \dd x \\ &=&\displaystyle -5 \cdot \cos(x) + \green C \end{array}$

$\int \cos(x) \; \dd x= \sin(x) + \green C$

Example

$\begin{array}{rcl} \displaystyle\int 3 \cdot \cos(x) \; \dd x &=&\displaystyle3 \cdot \int \cos(x) \; \dd x \\ &=&\displaystyle 3 \cdot \sin(x) + \green C \end{array}$

Determine an antiderivative $F$ of the function $f(x)=2\cdot \e^x$ .

$F(x)=$ $2\cdot \e^x$

$\begin{array}{rcl} \displaystyle \int f(x) \, \dd x&=&\displaystyle \int 2\cdot \e^x \, \dd x\\ &&\phantom{xxx}\blue{\text{substituted}}\\ &=&\displaystyle 2\cdot \int \e^x \, \dd x\\ &&\displaystyle\phantom{xxx}\blue{\text{rule of calculation }\int c \cdot f(x) \; \dd x=c\int f(x) \; \dd x}\\ &=&\displaystyle2\cdot \e^x+C\\ &&\displaystyle\displaystyle\phantom{xxx}\blue{\text{rule of calculation }\int \e^x \, \dd x = \e^x + C} \end{array}$

Since only one antiderivative is asked, we can choose $C=0$ . This gives
$F(x)=2\cdot \e^x$

New example