3 Integrals giving rise to logarithms

Example 21

Find ∫ 3 x 2 + 1 x 3 + x + 2 d x

Solution

Let us consider what happens when we make the substitution z = x 3 + x + 2 . Note that

d z d x = 3 x 2 + 1 so that we can write d z = ( 3 x 2 + 1 ) d x

Then

∫ 3 x 2 + 1 x 3 + x + 2 d x = ∫ 1 z d z = ln z + c = ln x 3 + x + 2

Note that in the last Example, the numerator of the integrand ( 3 x 2 + 1 ) is the derivative of the denominator ( x 3 + x + 2 ) . The result is the logarithm of the denominator. This is a special case of the following rule:

Key Point 7
∫ f ′ ( x ) f ( x ) d x = ln f ( x ) + c
Note that it is the modulus of f ( x ) in the answer.
Task!

Write down, purely by inspection, the following integrals:

  1. ∫ 1 x + 1 d x ,
  2. ∫ 2 x x 2 + 8 d x ,
  3. ∫ 1 x − 3 d x .

Hint: In each case the numerator of the integrand is the derivative of the denominator.

  1. ln | x + 1 | + c ,
  2. ln | x 2 + 8 | + c ,
  3. ln | x − 3 | + c
Task!

Evaluate the definite integral ∫ 2 4 3 t 2 + 2 t t 3 + t 2 + 1 d t .

ln t 3 + t 2 + 1 2 4 = ln 81 − ln 13 = 1.83

Sometimes it is necessary to make slight adjustments to the integrand to obtain a form for which the rule in Key Point 7 is suitable. Consider the next Example.

Example 22

Find the indefinite integral ∫ x 2 x 3 + 1 d x .

Solution

In this Example the derivative of the denominator is 3 x 2 whereas the numerator is just x 2 . We adjust the numerator as follows:

∫ x 2 x 3 + 1 d x = 1 3 ∫ 3 x 2 x 3 + 1 d x and integrate by the rule to get 1 3 ln x 3 + 1 + c

Note that the sort of procedure in the last Example is only possible because we can move constant factors through the integral sign. It would be wrong to try to move terms involving the variable x in a similar way.

Exercise

Write down the result of finding the following integrals.

  1. ∫ 1 x d x ,
  2. ∫ 2 t t 2 + 1 d t ,
  3. ∫ 1 2 x + 5 d x ,
  4. ∫ 2 3 x − 2 d x .
  1. ln | x | + c ,
  2. ln | t 2 + 1 | + c ,
  3. 1 2 ln | 2 x + 5 | + c ,
  4. 2 3 ln | 3 x − 2 | + c .