# A little guide to integration.

Common Integration Formulas

1. N k dx = kx + C

2. N xa dx = xa+1/(a+1) + C, a B -1

3. N ex dx = ex + C

4. N x-1 dx = ln (|x|) + C, x > 0

5. N k f(x) dx = k N f(x) dx

6. N [ f(x)±g(x) ] dx = N f(x) dx ± N g(x) dx

N.B.: Don[t forget adding an arbitrary constant C after integrating!

Examples:

N x4 dx = x5/5 + C

N 1/x dx = N x-1 dx = ln(|x|) + C

N (6/x + 6) dx = 6 N 1/x dx + N 6 dx = 6 ln(|x|) + 6x + C

N 5x dx = N ex ln(5) dx = 1/ln(5) ex ln(5) + C = 5x/ln(5) + C

But wait, I thought N ex dx = ex + C.

That is true, but don[t forget about the chain rule. Think about it backwards, using

differentiation.

d/dx 5x = 5x ln(5)

Therefore, d/dx 5x/ln(5) = 5x.

But is there an easier way of doing this? Yep.

www.notesolution.com

## Document Summary

: dont forget adding an arbitrary constant c after integrating! 1/x dx = x-1 dx = ln(|x|) + c. (6/x + 6) dx = 6 1/x dx + 6 dx = 6 ln(|x|) + 6x + c. 5x dx = ex ln(5) dx = 1/ln(5) ex ln(5) + c = 5x/ln(5) + c. But wait, i thought ex dx = ex + c. That is true, but dont forget about the chain rule. Think about it backwards, using differentiation. d/dx 5x = 5x ln(5) Let u = x ln(5) du = ln(5) dx. This is another method of solving the problem above, except in the example above, c=0. It can be used for simpler situations as well: Let u = x3, 1/3 du = x2 dx. But itd be wiser to just integrate that one directly. www. notesolution. com. Let u = ln(x2), du = 2/x dx.