Known antiderivatives of some quotient functions

Integration: Integration techniques

Known antiderivatives of some quotient functions

In the following pages, we will look at methods to find the antiderivative of some quotient functions. First, we'll look at two special antiderivatives.

\[\int \frac{1}{x^2+1} \; \dd x=\arctan(x)+\green C\]

Example

#\begin{array}{rcl}\displaystyle \int \frac{3}{x^2+1} \; \dd x &=& 3 \displaystyle \int\frac{1}{x^2+1} \; \dd x \\ &=& 3 \arctan(x) + \green C\end{array}#

Our first step is substituting #x=\tan(t)#, which gives:

\[\int \frac{1}{x^2+1} \; \dd x = \int \frac{1}{\tan^2(t)+1} \; \dd \tan(t)\]

We already know that #\frac{\dd \tan(t)}{\dd t} = \tan^2(t)+1#. Therefore, #\dd \tan(t) = \tan^2(t)+1 \; \dd t#. And thus:

\[\int \frac{1}{\tan^2(t)+1} \; \dd \tan(t)= \int \frac{\tan^2(t)+1}{\tan^2(t)+1} \; \dd t\]

We can solve this further. This goes as follows:

\[\begin{array}{rcl}\displaystyle \int \dfrac{\tan^2(t)+1}{\tan^2(t)+1} \; \dd t&=&\displaystyle \int 1 \; \dd t \\ &=&t+\green C\end{array}\]

We now substitute #\tan(t)=x#, in other words #t=\arctan(x)#, back again. This gives:

\[\int \frac{1}{x^2+1} \; \dd x = \arctan(x) +\green C\]

\[\int \frac{1}{\sqrt{1-x^2}} \; \dd x=\arcsin(x)+\green C\]

Example

#\begin{array}{rcl}\displaystyle \int \frac{2}{\sqrt{1-x^2}} \; \dd x &=& 2 \displaystyle \int\frac{1}{\sqrt{1-x^2}} \; \dd x \\ &=& 2 \arcsin(x) + \green C\end{array}#

Our first step is substituting #x=\sin(t)#, which gives:

\[\int \frac{1}{\sqrt{1-x^2}} \; \dd x = \int \frac{1}{\sqrt{1-\sin^2(t)}} \; \dd \sin(t)\]

We already know that : #\frac{\dd \sin(t)}{\dd t} = \cos(t)#. Therefore #\dd \sin(t) = \cos(t) \; \dd t#. And thus:

\[\int \frac{1}{\sqrt{1-\sin^2(t)}} \; \dd \sin(t)= \int \frac{\cos(t)}{\sqrt{1-\sin^2(t)}} \; \dd t\]

We can solve this further using #1-\sin^2(t)=\cos^2(t)#. This goes as follows:

\[\begin{array}{rcl}\displaystyle \int \frac{\cos(t)}{\sqrt{1-\sin^2(t)}} \; \dd t&=& \displaystyle \int \frac{\cos(t)}{\sqrt{\cos^2(t)}} \; \dd t \\ &=&\displaystyle \int \frac{\cos(t)}{|\cos(t)|} \; \dd t \end{array}\]

We now take #-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}# in which case #|\cos(t)|=\cos(t)# applies.

\[\begin{array}{rcl}\displaystyle \int \dfrac{\cos(t)}{|\cos(t)|} \; \dd t&=&\displaystyle \int \dfrac{\cos(t)}{\cos(t)} \; \dd t \\&=& \displaystyle \int 1 \; \dd t\\ &=&t+\green C\end{array}\]

We now substitute #\sin(t)=x# with #-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}#, in other words #t=\arcsin(x)#, back again. This gives:

\[\int \frac{1}{\sqrt{1-x^2}} \; \dd x = \arcsin(x) +\green C\]

Determine #\int {{1}\over{4\cdot x^2+1}} \,\dd x#.

Use #C# to denote the constant of integration.

#\int {{1}\over{4\cdot x^2+1}} \,\dd x=# #{{\arctan(2\cdot x)}\over{2}} + C#

#\begin{array}{rcl}\displaystyle \int {{1}\over{4\cdot x^2+1}} \; \dd x &=& \displaystyle \int \frac{1}{2} \frac{1}{(2\cdot x)^2+1} \; \dd(2\cdot x) \\&&\phantom{xxx}\blue{\text{rewritten so we can substitute }} \\ &=& \displaystyle \int \frac{1}{2} \cdot {{1}\over{u^2+1}} \; \dd u \\&&\phantom{xxx}\blue{\text{substituting }2\cdot x=u} \\ &=& \frac{1}{2} \cdot \arctan(u) +C\\&&\phantom{xxx}\blue{\text{found the antiderivative}} \\ &=&\displaystyle {{\arctan(2\cdot x)}\over{2}} +C \\&&\phantom{xxx}\blue{\text{substituting }u=2\cdot x} \end{array}#

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