# 1. Theory

Consider an integral of the form:

ƒ(ax + b)

where a and b are constants. We have here an unspecified function ƒ of a linear function x.

Letting u = ax + b, then du/dx = a, and this gives dx = du/a

This allows us to change the integration variable from x to u.

ƒ(ax + b) dx =   ƒ(u)du/a

The final result is

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where u = ux + b

This is a general result for integrating functions of a linear function x.

Each application of this result involves dividing by the coefficient of x and then integrating.

# 2. Exercises

Perform the following integrations.

Click on the questions to reveal their solution

Exercise 1:

(2x − 1)3dx

Solution: Let u = 2x − 1, then du/dx = 2 and dx = du/2.

 ∫ (2x−1)3dx =  ∫ u3 du/2 = 1/2 ∫ u3du = 1/2⋅1/4 u4+C = 1/8 (2x − 1)4 + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 2 and   ƒ(u) du is   u3du.

Exercise 2:

cos (3x + 5) dx

Solution: Let u = 3x + 5, then du/dx = 3 and dx = du/3.

 ∫ cos (3x + 5) dx =  ∫ cos (u)⋅ du/3 = 1/3 ∫ cos (u) du = 1/3 sin (u) + C = 1/3 sin (3x + 5)4 + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 3 and   ƒ(u) du is   cos (u) du.

Exercise 3:

e5x + 2dx

Solution: Let u = 5x + 2, then du/dx = 5 and dx = du/5.

 ∫ e5x+2dx =  ∫ eu du/5 = 1/5 ∫ eudu = 1/5 eu + C = 1/5 e5x + 2 + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 5 and   ƒ(u) du is   eudu.

Exercise 4:

sinh (3x) dx

Solution: Let u = 3x, then du/dx = 3 and dx = du/3.

 ∫ sinh (3x) dx =  ∫ sinh (u)⋅du/3 = 1/3 ∫ sinh (u) du = 1/3 cosh (u) + C = 1/3 cosh (3x) + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 3 and   ƒ(u) du is   sinh (u) du.

Exercise 5:

dx2x − 1

Solution: Let u = 2x − 1, then du/dx = 2 and dx = du/2.

 ∫ dx2x − 1 =  ∫ duu⋅12 = 1/2 ∫ duu = 1/2 ln u + C = 1/2 ln 2x − 1 + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 2 and   ƒ(u) du is   du/u

Exercise 6:

dx1 + (5x)2

Solution: Let u = 5x − 1, then du/dx = 5 and dx = du/5.

 ∫ dx1 + (5x)2 =  ∫ du1 + u2⋅15 = 1/5 ∫ du1 + u2 = 1/5 tan−1 u + C = 1/5 tan−1 5x + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 5 and   ƒ(u) du is   du/1 + u2

Exercise 7:

sec2 (7x + 1) dx

Solution: Let u = 7x + 1, then du/dx = 7 and dx = du/7.

 ∫ sec2 (7x + 1) dx = 1/7 ∫ sec2 u du = 1/7 tanu + C = 1/7 tan (7x + 1) + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 7 and   ƒ(u) du is   sec2udu.

Exercise 8:

sin (3x − 1) dx

Solution: Let u = 3x − 1, then du/dx = 3 and dx = du/3.

 ∫ sin (3x − 1) dx = 1/3 ∫ sin (u) du = −1/3 cos (u) + C = −1/3 cos (3x − 1) + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 3 and   ƒ(u) du is   sin (u) du.

Exercise 9:

cosh (1 + 2x) dx

Solution: Let u = 1 + 2x, then du/dx = 2 and dx = du/2.

 ∫ cosh (1 + 2x) dx = = 1/2 ∫  cosh (u) du = 1/2 sinh (u) + C = 1/2 sinh (1 + 2x) + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 2 and   ƒ(u) du is   cosh (u) du.

Exercise 10:

tan (9x − 1) dx

Solution: Let u = 9x − 1, then du/dx = 9 and dx = du/9.

 ∫ tan (9x − 1) dx =  ∫ tan(u) du/9 = 1/9 ∫ tan (u) du = −1/9 ln cos (u) + C = −1/9 ln cos (9x − 1) + C

Note: The final result can also be obtained using the general pattern:

ƒ(ax + b) dx = 1/a  ƒ(u) du

- where a = 9 and   ƒ(u) du is   tan (u) du.

# 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)