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In what follows, c is a constant of integration and can take any contant value. You may want to use a the integral table and properties in "table of integrals and a table of properties of integrals".
Example 1: Evaluate the integral
ò ln(2x + 1) dx
Solution to Example 1:
Substitution: Let u = 2x + 1 which leads to du / dx = 2
or du = 2 dx or dx = du / 2, the above integral becomes
ò ln(2x + 1) dx = (1/2) ò ln u du
We now use integral formulas for ln x function to obtain
ò ln(2x + 1) dx = (1 / 2) [u ln u - u] + c
We now substitute u by 2x + 1 into the above to obtain
ò ln(2x + 1) dx = (1 / 2)(2x + 1) ln (2x + 1) - (1 / 2)(2x + 1) + C
= (1 / 2)(2x + 1) ln (2x + 1) - x - 1/2 + C
= (1 / 2)(2x + 1) ln (2x + 1) - x + k , where k = c - 1/2 and is a constant.
Check: Differentiate (1 / 2)(2x + 1) ln (2x + 1) - x + k and see that you obtain ln(2x + 1) which is the integrand in the given integral. This is a way to check the answer to indefinite integrals evaluation.
Example 2: Evaluate the integral
ò x ln x dx
Solution to Example 2:
Let f(x) = ln x and g ' (x) = x
which gives f'(x) = 1 / x and g(x) = x 2 / 2.
Using the integration by parts
ò f(x) g '(x) dx = f(x) g(x) - òf '(x) g(x) dx , we obtain
ò x ln x dx = [ x 2 / 2] ln x - ò [ x 2 / 2 ] [1 / x] dx
= [ x 2 / 2] ln x - ò [ x / 2 ] dx =
= [ x 2 / 2] ln x - x 2 / 4 + c.
Practice: Differentiate [ x 2 / 2] ln x - x 2 / 4 + c to obtain the integrand x ln x in the given integral.
Example 3: Evaluate the integral
ò ln(x) / x dx
Solution to Example 3:
Let u = ln x so that du / dx = 1 / x; the given integral can be written as
= ò u du
Integrate to obtain
= u 2 / 2 + c
Substitute u by ln x
= [ln x] 2 / 2 + c
As an exercise, check the final answer by differentiation.
Exercises: Evaluate the following integral.
1. ò x 3 ln x dx
2. ò [x - ln x] dx
Answers to Above Exercises
1. x 4 ln x / 4 - x 4 / 16 + c
2. -x ln x + x 2 / 2 + x + c
More references on
integrals and their applications in calculus.
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