f(u) can take on many forms. Several of these forms are presented in the following examples. dx, x is a variable.

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1 MATH 56: INTEGRATION USING u-du SUBSTITUTION: u-substitution and the Indefinite Integral: An antiderivative of a function f is a function F such that F (x) = f (x). Any two antiderivatives of f differ at most by a constant. The most general antiderivative of f is called the indefinite integral of f and is denoted by f (x)dx = F(x) + C, where C is called the constant of integration. Some basic integration formulas are as follows:. k dx = kx + C, k is a constant. x n dx = xn+ +C, n - 3. e x dx = e x +C 4. k f (x)dx = k f(x)dx, k is a constant 5. [ f (x) ± g(x)]dx = f (x)dx ± g(x)dx 6. dx = ln x +C x The above integrals can only be used when x is a variable. This section is to explain cases where u = f (x) and u is a differentiable function of x. The method of u-substitution is used to reverse derivatives that were obtained using the chain rule. These integrals have the following form: fg(x) ( ) g (x)dx = f (u)du, where u = g(x) and du = g (x)dx. f(u) can take on many forms. Several of these forms are presented in the following examples. Power Rule for Integration: u n du = un+ + C, if n - IMPORTANT: The difference between the formula for x n dx and the power rule for integration is that in the power rule, u represents a function and in x n dx, x is a variable. Example : Evaluate (x +) 0 dx. Solution: Since the integrand is a power of the function (x + ), we let u = x +. We then let du = () dx (since is the derivative of x + ). Now substitute these values into your original equation. (x +) 0 dx = u 0 du = u (x +) + C = + C

2 Integration of this form is often known as integration by substitution (or u-du substitution). Note that the above example uses the following general steps: To evaluate an indefinite integral by u-du substitution: Step : Rewrite your problem algebraically, if necessary, to simplify, expand, remove radicals, eliminate denominators, etc. Step : Choose u Step 3: Find du. Step 4: Make a constant adjustment, if necessary (see below in Example 3). Step 5: Rewrite original integral completely in terms of u and du. Make sure that you don t have both u s and x s in your problem after this step. Step 6: Apply appropriate rule to evaluate the integral (don t forget to add + C at the end) Step 7: Now rewrite your answer in terms of the original variable. Example : Evaluate 3x (x 3 +7) 3 dx Solution: Step : No need to rewrite the problem algebraically. Step : Choose u: The integrand contains a power of the function (x 3 + 7), so let u = x Step 3: Find du.: du = 3x dx Step 4: Not necessary. Step 5: Rewrite using u s and du s: 3x (x 3 +7) 3 dx = (x 3 +7) 3 (3x dx) = u 3 du Step 6: Apply the Power Rule for Integration: u 3 du = u4 4 +C Step 7: Rewrite in terms of the original variable u C = (x3 + 7) 4 + C 4 Sometimes, using simple u-substitution (as in the previous two examples) is not enough. Sometimes a constant adjustment must be made in order to obtain du in the integrand, as Example 3 illustrates. Example 3: Evaluate x x + 5 dx Solution: Step : x x + 5 dx = x(x + 5) dx Step : u = x + 5 Step 3: du = xdx

3 this Step 4: Since the problem consists of (x + 5) (xdx), du is not in the integrand and integral does not have the form u n du. We can put the integral in this form, however, by first multiplying and then dividing the integral by. This does not change the value of the integral. x(x + 5) dx = x(x + 5) dx = (x + 5) [xdx] = (x + 5) [xdx] Step 5: x(x + 5) dx = (x + 5) [xdx] = u du Step 6: u du = u 3 + C = 3 3 u 3 + C Step 7: 3 u 3 +C = 3 (x + 5) 3 +C Note that we can multiply the integrand by a non-zero constant k as long as we compensate for this by multiplying the integral by /k. IMPORTANT: We can adjust for constant factors by multiplying and dividing as described above, but we cannot adjust for variable factors this way! The following is an example of a problem where the Power Rule for Integrals does not apply: Example 4: Evaluate 4x (x 4 +) dx Solution: If we set u = x 4 +, we get du = 4x 3 dx. To get du in the integral, we need an additional factor of the variable x. However, we can only adjust for constant factors. Thus we cannot use the power rule. To evaluate the integral, we must first expand (x 4 + ) and then multiply through by 4x. Then integrate the results of this algebraic operation. The resulting solution follows (work is not shown. The solution is left for you to verify): 4x (x 4 +) dx = 4 x + x7 7 + x3 + C 3 Integrating Natural Exponential Functions: e u du = e u +C Example 5: Evaluate xe x dx Solution:. No need to rewrite algebraically.. u = x 3. du = xdx

4 4. Not necessary to make a constant adjustment. 5. xe x dx = e x [x dx ] = e u du 6. e u du = e u +C 7. e u + C = e x + C

5 Integrating Functions involving /u du: du = ln u +C u Example 6: (x 3 + 3x) Evaluate x 4 + 3x dx + 7 (x 3 + 3x) Solution:. x 4 + 3x + 7 dx = (x 3 + 3x) (x 4 + 3x + 7) dx. u = x 4 + 3x du = (4x 3 + 6x)dx (Note that this is two times the numerator) 4. (x 3 + 3x) (x 4 + 3x + 7) dx = (x4 + 3x + 7) [(x 3 + 3x)]dx 5. (x4 + 3x + 7) [(4x 3 + 6x)dx ] = du u 6. du = ln u +C u 7. ln x4 + 3x C or ln x 4 + 3x + 7 +C = (x4 + 3x + 7) [(4x 3 + 6x)dx ] Also note that some integrals may consist of a sum of terms where each term has a different form. Just remember to integrate each term separately, applying the appropriate rules to each term. Basic Integration Formulas discussed so far: If u is a differentiable function of x:. k du = ku + C, k is a constant. u n du = un+ +C, n - 3. e u du = e u +C 4. k f (x)dx = k f(x)dx, k is a constant 5. [ f (x) ± g(x)]dx = f (x)dx ± g(x)dx 6. du = ln u +C u