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Integration formulas
Differentiation Formula Indefinite Integral
d d x ( k ) = 0 k d x = k x 0 d x = k x + C
d d x ( x n ) = n x n 1 x n d n = x n + 1 n + 1 + C for n 1
d d x ( ln | x | ) = 1 x 1 x d x = ln | x | + C
d d x ( e x ) = e x e x d x = e x + C
d d x ( sin x ) = cos x cos x d x = sin x + C
d d x ( cos x ) = sin x sin x d x = cos x + C
d d x ( tan x ) = sec 2 x sec 2 x d x = tan x + C
d d x ( csc x ) = csc x cot x csc x cot x d x = csc x + C
d d x ( sec x ) = sec x tan x sec x tan x d x = sec x + C
d d x ( cot x ) = csc 2 x csc 2 x d x = cot x + C
d d x ( sin −1 x ) = 1 1 x 2 1 1 x 2 = sin −1 x + C
d d x ( tan −1 x ) = 1 1 + x 2 1 1 + x 2 d x = tan −1 x + C
d d x ( sec −1 | x | ) = 1 x x 2 1 1 x x 2 1 d x = sec −1 | x | + C

From the definition of indefinite integral of f , we know

f ( x ) d x = F ( x ) + C

if and only if F is an antiderivative of f . Therefore, when claiming that

f ( x ) d x = F ( x ) + C

it is important to check whether this statement is correct by verifying that F ( x ) = f ( x ) .

Verifying an indefinite integral

Each of the following statements is of the form f ( x ) d x = F ( x ) + C . Verify that each statement is correct by showing that F ( x ) = f ( x ) .

  1. ( x + e x ) d x = x 2 2 + e x + C
  2. x e x d x = x e x e x + C
  1. Since
    d d x ( x 2 2 + e x + C ) = x + e x ,

    the statement
    ( x + e x ) d x = x 2 2 + e x + C

    is correct.
    Note that we are verifying an indefinite integral for a sum. Furthermore, x 2 2 and e x are antiderivatives of x and e x , respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section.
  2. Using the product rule, we see that
    d d x ( x e x e x + C ) = e x + x e x e x = x e x .

    Therefore, the statement
    x e x d x = x e x e x + C

    is correct.
    Note that we are verifying an indefinite integral for a product. The antiderivative x e x e x is not a product of the antiderivatives. Furthermore, the product of antiderivatives, x 2 e x / 2 is not an antiderivative of x e x since
    d d x ( x 2 e x 2 ) = x e x + x 2 e x 2 x e x .

    In general, the product of antiderivatives is not an antiderivative of a product.
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Verify that x cos x d x = x sin x + cos x + C .

d d x ( x sin x + cos x + C ) = sin x + x cos x sin x = x cos x

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In [link] , we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum f + g . In [link] a. we showed that an antiderivative of the sum x + e x is given by the sum ( x 2 2 ) + e x —that is, an antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if F and G are antiderivatives of any functions f and g , respectively, then

d d x ( F ( x ) + G ( x ) ) = F ( x ) + G ( x ) = f ( x ) + g ( x ) .

Therefore, F ( x ) + G ( x ) is an antiderivative of f ( x ) + g ( x ) and we have

( f ( x ) + g ( x ) ) d x = F ( x ) + G ( x ) + C .

Similarly,

( f ( x ) g ( x ) ) d x = F ( x ) G ( x ) + C .

In addition, consider the task of finding an antiderivative of k f ( x ) , where k is any real number. Since

d d x ( k f ( x ) ) = k d d x F ( x ) = k F ( x )

for any real number k , we conclude that

k f ( x ) d x = k F ( x ) + C .

These properties are summarized next.

Properties of indefinite integrals

Let F and G be antiderivatives of f and g , respectively, and let k be any real number.

Sums and Differences

( f ( x ) ± g ( x ) ) d x = F ( x ) ± G ( x ) + C

Constant Multiples

k f ( x ) d x = k F ( x ) + C

From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see [link] b. for an example involving an antiderivative of a product.) We look at and address integrals involving these more complicated functions in Introduction to Integration . In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

Practice Key Terms 3

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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