# 1. Theory

Consider an integral of the form:

ƒ′(x)/ƒ(x)dx

Letting u = ƒ(x), then du/dx = ƒ′(x), and this gives: du = ƒ′(x) dx.

So:

 ∫  ƒ′(x)/ƒ(x) dx = ∫  1/ƒ(x)⋅ƒ′(x) dx = ∫  du/u = ln u + C

The general result is:

 ∫  ƒ′(x)/ƒ(x)dx = ln ƒ(x) + C

This is a useful generalisation of the standard integral:

1/xdx = ln x + C

that applies straightforwardly when the top line of what we are integrating is a multiple of the derivative of the bottom line.

# 2. Exercises

Perform the following integrations.

Click on the questions to reveal their solution

Exercise 1:

10x + 3/5x2 + 3x − 1dx

Solution:

10x + 3/5x2 + 3x − 1 dx is of the form     ƒ′(x)/ƒ(x) = ln ƒ(x) + C

Let u = 5x2 + 3x − 1 then du/dx = 10x + 3 and du = (10x + 3) dx

 ∴     ∫  10x + 3/5x2 + 3x − 1 dx = ∫  du/u = ln u + C = ln 5x2 + 3x − 1 + C

Exercise 2:

3x2/x3 − 6dx

Solution:

3x2/x3−6 dx is of the form     ƒ′(x)/ƒ(x) = ln ƒ(x) + C

Let u = x3 − 6 then du/dx = 3x2 and du = (3x2) dx

 ∴     ∫  3x2/x3 − 6 dx = ∫  1/u du = ln u + C = ln x3 − 6 + C

Exercise 3:

4x2/x3 − 1 dx

Solution:

4x2/x3−1 dx is of the slightly more general form     k ƒ′(x)/ƒ(x) = k     ƒ′(x)/ƒ(x) = kln ƒ(x) + C

where k is a constant, i.e. the top line is a multiple of the derivative of the bottom line.

Let u = x3 − 1 then du/dx = 3x2 and du/3 = (x2) dx

 ∴   4   ∫  x2/x3−1 dx = 4   ∫  1/u du/3 = 4/3ln u + C = 4/3ln x3−1 + C

Exercise 4:

2x − 1/x2x − 7dx

Solution:

4x2/x3−1 dx is of the form     ƒ′(x)/ƒ(x) = ln ƒ(x) + C

Let u = x2x − 7 then du/dx = 2x − 1 and du = (2x − 1) dx

 ∴     ∫  2x − 1/x2 − x − 7 dx = ∫  du/u = ln u + C = ln x2 − x − 7 + C

Exercise 5:

x − 1/x2 − 2x + 7dx

Solution:

x − 1/x2 − 2x + 7 dx is of the slightly more general form     kƒ′(x)/ƒ(x) = k    ƒ′(x)/ƒ(x) = k ln ƒ(x) + C

where k is a constant, i.e. the top line is a multiple of the derivative of the bottom line.

Let u = x2 − 2x + 7, then du/dx = 2x − 2 and du = (x2) dx

so that du/2 = (x − 1) dx

 ∴  4  ∫  x − 1/x2 − 2x + 7 dx = ∫  1/2 du/u = 1/2 ln u + C = 1/2 ln x2 − 2x + 7 + C

Exercise 6:

8x/x2 − 4dx

Solution:

Let u = x2 − 4 then du/dx = 2x and du/2 = (x) dx

 ∴     ∫  8x/x2 − 4 dx = 8  ∫  1/u du/2 = 4 ln u + C = 4 ln x2 − 4 + C

Exercise 7:

x2/x3 + 2 dx

Solution:

Let u = x3 + 2 then du/dx = 3x2 and du/3 = (x2) dx

 ∴     ∫  x2/x3 + 2 dx = ∫  1/u du/3 = 1/3 ln u + C = 1/3 ln x3 + 2 + C

Exercise 8:

sin (x)/cos (x)dx

Solution:

Let u = cos (x) then du/dx = − sin (x) and du/(−1) = sin (x) dx

 ∴     ∫  sin (x)/cos (x) dx = ∫  1/u du/(−1) = −  ∫  du/u = − ln u + C = − ln cos (x) + C

Exercise 9:

cos (x)/sin (x) dx

Solution:

Let u = sin (x) then du/dx = cos (x) and du= cos (x) dx

 ∴     ∫  cos (x)/sin (x) dx = ∫  1/u du = ln u + C = ln sin (x) + C

Exercise 10:

cos (x)/5 + sin (x) dx

Solution:

Let u = 5 + sin (x) then du/dx = cos (x) and du= cos (x) dx

 ∴     ∫  cos (x)/5 + sin (x) dx = ∫  1/u du = ln u + C = ln 5 + sin (x) + C

Exercise 11:

sinh (x)/1−cosh (x) dx

Solution:

Let u = 1 − cosh (x) then du/dx = −sinh (x) and du= sinh (x) dx

 ∴     ∫  sinh (x)/1 − cosh (x) dx = ∫  1/u (−du) = − ln u + C = − ln 1 − cosh (x) + C

# 3. Standard Integrals

ƒ(x) ∫ƒ(x) dx ƒ(x) ∫ƒ(x) dx xn xn+1/n+1 (n ≠ −1) [g(x)]n g'(x) [g(x)]n+1/n+1 (n ≠ −1) 1/x ln x g'(x)/g(x) ln g(x) ex ex ax ax/ln a (a > 0) sin x −cos x sinh x cosh x cos x sin x cosh x sinh x tan x − ln cosx tanh x ln cosh x cosec x ln tan  x/2 cosech x ln tanh  x/2 sec x ln sec x + tan x sec x 2 tan−1ex sec2 x tan x sec2 x tanh x cot x ln sin x cot x ln sinh x sin2 x x/2 − sin 2x/4 sinh2 x sinh 2x/4 − x/2 cos2 x x/2 + sin 2x/4 cosh2 x sinh 2x/4 + x/2 1/a2 + x2 1/a tan −1  x/a   (a > 0) √a2 + x2 a2/2[ sinh−1(x/a) + x √a2−x2/a2] 1/a2 − x2 1/2a ln  a + x/a − x    (0 < x < a) √a2−x2 a2/2[ sin−1(x/a) + x √a2−x2/a2] 1/x2 − a2 1/2a ln  x − a/x + a   (x > a > 0) √x2 − a2 a2/2[−cosh−1(x/a) + x √x2−a2/a2] 1/ √ a2 + x2 ln  x + √a2 + x2/a   (a > 0) 1/ √ a2 − x2 sin−1  x/a     (−a < x < a) 1/ √ x2 − a2 ln  x + √x2 − a2/a   (x > a > 0)