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Integration

To integrate: Add one to the power and divide by the new power. When integrating always add a constant,  
\[c\]
.
\[4x^3\]
  integrated is  
\[\int 4 x^{3}dx=\frac{4x^{3+1}}{3+1}+c = x^4+c\]
.
The  
\[\int\]
  symbol means integrate and the  
\[dx\]
  above means integrate with respect to  
\[x\]
.
We can integrate a sum using the same rule for each term.
\[ 2x^5-4x^7\]
  when integrated is  
\[\int 2x^5-4x^7 dx = \frac{2x^{5+1}}{5+1}\frac{4x^{7+1}}{7+1}+c = \frac{x^6}{3}-\frac{x^8}{2}+c\]
.
This rule 'add one to the power and divide by the new power' works for  
\[x\]
's and constants too.
To integrate  
\[3x\]
  write as  
\[3x^1\]
  then apply the rule to give  
\[\frac 3x^1 dx = \frac{3x^{1+1}}{1+1}+c = \frac{3x^2}{2}+c\]
.
To integrate  
\[4\]
  write as  
\[4x^0\]
  then integrate using the above gives  
\[\int 4x^0 dx = \frac{4x^{0+1}}{0+1}+c=4x+c\]
.
Integrate  
\[4x^2-6x-4\]
.
Write  
\[4x^2-6x^1-4x^0\]
.
We have
\[\begin{equation} \begin{aligned} \int 4x^2-6x^1-4x^0 dx &= \frac{4x^{2+1}}{2+1}- \frac{6x^{1+1}}{1+1}- \frac{4x^{0+1}}{0+1}+c \\ &= \frac{4x^3}{3}-3x^2-4x+c \end{aligned} \end{equation}\]

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