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}#
\[\int \frac{1}{\sqrt{1-x^2}} \; \dd x=\arcsin(x)+\green C\]
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}#
#\int {{1}\over{8\cdot x^2+1}} \,\dd x=# #{{\arctan \left(2^{{{3}\over{2}}}\cdot x\right)}\over{2^{{{3}\over{2}}}}} + C#
#\begin{array}{rcl}\displaystyle \int {{1}\over{8\cdot x^2+1}} \; \dd x &=& \displaystyle \int \frac{1}{2^{{{3}\over{2}}}} \frac{1}{(2^{{{3}\over{2}}}\cdot x)^2+1} \; \dd(2^{{{3}\over{2}}}\cdot x) \\&&\phantom{xxx}\blue{\text{rewritten so we can substitute }} \\ &=& \displaystyle \int \frac{1}{2^{{{3}\over{2}}}} \cdot {{1}\over{u^2+1}} \; \dd u \\&&\phantom{xxx}\blue{\text{substituting }2^{{{3}\over{2}}}\cdot x=u} \\ &=& \frac{1}{2^{{{3}\over{2}}}} \cdot \arctan \left(u\right) +C\\&&\phantom{xxx}\blue{\text{found the antiderivative}} \\ &=&\displaystyle {{\arctan \left(2^{{{3}\over{2}}}\cdot x\right)}\over{2^{{{3}\over{2}}}}} +C \\&&\phantom{xxx}\blue{\text{substituting }u=2^{{{3}\over{2}}}\cdot x} \end{array}#
#\begin{array}{rcl}\displaystyle \int {{1}\over{8\cdot x^2+1}} \; \dd x &=& \displaystyle \int \frac{1}{2^{{{3}\over{2}}}} \frac{1}{(2^{{{3}\over{2}}}\cdot x)^2+1} \; \dd(2^{{{3}\over{2}}}\cdot x) \\&&\phantom{xxx}\blue{\text{rewritten so we can substitute }} \\ &=& \displaystyle \int \frac{1}{2^{{{3}\over{2}}}} \cdot {{1}\over{u^2+1}} \; \dd u \\&&\phantom{xxx}\blue{\text{substituting }2^{{{3}\over{2}}}\cdot x=u} \\ &=& \frac{1}{2^{{{3}\over{2}}}} \cdot \arctan \left(u\right) +C\\&&\phantom{xxx}\blue{\text{found the antiderivative}} \\ &=&\displaystyle {{\arctan \left(2^{{{3}\over{2}}}\cdot x\right)}\over{2^{{{3}\over{2}}}}} +C \\&&\phantom{xxx}\blue{\text{substituting }u=2^{{{3}\over{2}}}\cdot x} \end{array}#
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