 | ( | 4x |  |  2 | + | 3x 1.1 6) | dx |
Look at the given expression,
| ( | 4x | | 2 | + | 3x1.1 6) | dx |
The first rule to remember is this: constant coefficients (such as the 4 in front of the x, the (1/2) in front of the x2, and the 3 in front of the x1.1) always "go along for the ride." In other words, we can just copy them over when taking the antiderivative. Therefore, the first term is
 | 4x dx | = | 4 | | x dx | = | 4 | 2 | + C | = | 2x2 + C. |
Now look at the second term.
| 2 | dx |
is the same as
| 1 2 | x2 | dx | = | 1 2 | | x2 | dx | = | 2 |
3 |
+ C | = | 6 |
+ C. |
That takes care of the second term. The third term is
| 3x1.1 dx | = | 3 | | x1.1 dx | = | 3 | 1.1+1) 1.1+1) | + C | = | 3x0.1 0.1 | + C. |
Finally, the last term is
| 6 dx | = | 6x + C. |
Putting them all together gives:
| ( | 4x | | 2 | + | 3x1.1 6) | dx |
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