### Integration: Antiderivatives

### Antiderivatives and the chain rule

We have seen the *chain rule* when we differentiated. This can also play a role in integrating functions. We will first look at how we integrate functions of the form #f(ax+b)# using the chain rule. Later, we'll discuss the *substitution method*, which is a more general form of integration with the consequences of the chain rule.

A function #f(\blue ax+b)# has the function #\frac{1}{\blue a}F(\blue ax+b)+\green C# as an antiderivative, in which #F# is an antiderivative of #f#. Therefore:

\[\displaystyle \int f(\blue ax+b) \; \dd x = \frac{1}{\blue a} F(\blue ax+b)+\green C\]

**Example**

#\begin{array}{rcl}\displaystyle \int \sin(\blue 3 x+5) \; \dd x&=&\frac{1}{\blue 3} \cdot -\cos(\blue3 x+5) \\ &=& -\frac{1}{3}\cos(3x+5)\end{array}#

#F(x)=# #{{\left(4\cdot x-3\right)^5}\over{20}}#

We use the chain rule for integration to determine the integral, and we note that the function is of the form #g(4\cdot x -3)#, with #g(x)=x^{4}#.

\[\begin{array}{rcl}

\displaystyle \int \left(4\cdot x-3\right)^4 \, \dd x &=&\displaystyle \int g(4 \cdot x -3) \, \dd x\\

&&\phantom{xxx}\blue{\text{rewritten to the form }g(ax+b) \text{ with } g(x)=x^{4}}\\

&=&\displaystyle \frac{1}{4}\cdot G(4 \cdot x -3)\\

&&\phantom{xxx}\blue{\text{chain rule for integration}}\\

&=&\displaystyle \frac{1}{4}\cdot \frac{1}{4+1}(4 \cdot x -3)^{4+1}+C\\

&&\displaystyle \phantom{xxx}\blue{\text{use rule } \int x^{n} \; \dd x =\frac{1}{n+1} x^{n+1} + C}\\

&=&\displaystyle {{\left(4\cdot x-3\right)^5}\over{20}}+C\\

&&\phantom{xxx}\blue{\text{simplified}}

\end{array}\]

Because we only asked for one antiderivative, we can now choose #C=0#. This gives:

\[F(x)={{\left(4\cdot x-3\right)^5}\over{20}}\]

We use the chain rule for integration to determine the integral, and we note that the function is of the form #g(4\cdot x -3)#, with #g(x)=x^{4}#.

\[\begin{array}{rcl}

\displaystyle \int \left(4\cdot x-3\right)^4 \, \dd x &=&\displaystyle \int g(4 \cdot x -3) \, \dd x\\

&&\phantom{xxx}\blue{\text{rewritten to the form }g(ax+b) \text{ with } g(x)=x^{4}}\\

&=&\displaystyle \frac{1}{4}\cdot G(4 \cdot x -3)\\

&&\phantom{xxx}\blue{\text{chain rule for integration}}\\

&=&\displaystyle \frac{1}{4}\cdot \frac{1}{4+1}(4 \cdot x -3)^{4+1}+C\\

&&\displaystyle \phantom{xxx}\blue{\text{use rule } \int x^{n} \; \dd x =\frac{1}{n+1} x^{n+1} + C}\\

&=&\displaystyle {{\left(4\cdot x-3\right)^5}\over{20}}+C\\

&&\phantom{xxx}\blue{\text{simplified}}

\end{array}\]

Because we only asked for one antiderivative, we can now choose #C=0#. This gives:

\[F(x)={{\left(4\cdot x-3\right)^5}\over{20}}\]

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