## Integration Rules

THEOREM 2

Let u and v be functions of x whose domains are an open interval I and suppose du and dv exist for every x in I.

 (i) du = u + C. (ii) Constant Rule c du = c du. (iii) Sum Rule du + dv = du + dv. (iv) Power Rule where r is rational, r -1,and u > 0 on I. (v) sin u du = -cos u + C. (vi) cos u du = sin u + C. (vii) eu du = eu 4- C. (viii) 1/u du = ln |u| + C (u 0).

Discussion

The Power Rule gives the integral of ur when r -1, while Rule (viii) gives the integral of ur when r = -1. When we put u =f(x) and v = g(x), the Constant and Sum Rules take the form

Constant Rule

c f(x) dx = c f(x) dx.

Sum Rule

(f (x) + g(x)) dx = f (x) dx + g(x) dx.

In the Constant and Sum Rules we are multiplying a family of functions by a constant and adding two families of functions. If we do either of these two things to families of functions differing only by a constant, we get another family of functions differing only by a constant. For example,

7(3x4 + C) = 21x4 + 7C = 21x4 + C'

is the family of all functions equal to 21x4 plus a constant. Similarly,

(3√x + C) + (5x - √x + D) = 5x + 2√x + (C + D) = 5x + 2√x + C

is the family of all functions equal to 5x + 2√x plus a constant.

PROOF OF THEOREM 2

 (i) This is just a short form of the theorem that u + C is the family of all functions which have the same derivative as u. (ii) We have c du = d(cu), whence c du = d(cu) = cu + C = c(u + C') = cdu. (iii) du + dv = d(u + v), du + dv = d(u + v) = u + v + C = du + du. (iv)

Rules (v)-(viii) are similar. Only the last formula, (viii), requires an explanation. The absolute value in ln |u| comes about by combining the two cases u > 0 and u < 0. When u > 0, u = |u| and

d(ln |u|) = d(ln u) = -du.

When w < 0, ln u is undefined, but |u| = - u and ln |u| = ln (-u). Thus

d(ln|u|) = d(ln(-u)) = -1/u d(-u) = -1/u du.

Thus, in both cases, when u 0,

Last Update: 2006-12-01