(1.) ∫ dx/(2x - 5)

= (1/2) ln (2x - 5) + C integrate

(2.) ∫dx/(5x + 10)

= (1/5)ln (5x + 10) + C integrate

(3.) ∫dx/(3x + 11)

= (1/3)ln (3x + 11) + C integrate

(4.) ∫dx/(3x - 11)

= (1/3)ln (3x - 11) + C integrate

(5.) ∫dx/(x² + 9)

= (1/9)∫dx/[(x²/9) + 1] factor like this

Let u = (x/3)

du = (1/3)dx take the derivative

3du = dx multiply each side by 3, cancel

= (1/3)∫du/(u² + 1) make substitutions

= (1/3) arctan u + C integrate

= (1/3) arctan (x/3) + C replace u with x/3

(6.) ∫dx/(2x² + 8) here is the problem

= (1/2)∫dx/(x² + 4) factor

= (1/2)(1/4)∫dx/[(x/2)² + 1] factor 4 out front on bottom

Let u = x/2

du = (1/2)dx take derivatives

2 du = dx multiply each side by 2, cancel

= (1/4)∫du/(u² + 1) make substitutions

= (1/4) arctan u + C integrate

= (1/4) arctan (x/2) + C replace u with x/2

(7.) ∫dx/(x² - 16)

= ∫dx/(x - 4)(x + 4) factor

A B use partial fractions
_______ + _______
x - 4 x + 4

Ax + 4A + Bx - 4B = 1 here is the equation

A + B = 0

4A - 4B = 1 here is the system

4A + 4B = 0 multiply thru by 4
_____________
8A = 1 add equations

A = 1/8 divide each side by 8, cancel

(1/8) + B = 0 replace A with 1/8

-1/8 -1/8 subtract 1/8 from each side
_________________
B =-1/8

(1/8)∫dx/(x - 4) - (1/8)∫dx/(x + 4) make substitutions

= (1/8)ln (x - 4) - (1/8)ln (x + 4) + C integrate

(8.) ∫ dx/(x² - 12) here is the problem

= ∫ A B
____________ + ______________
x + 12^1/2 x - 12^1/2 use partial fractions

Ax - A(12)^1/2 + Bx + B(12)^1/2 = 1 here is the equation

A + B = 0

-A(12)^1/2 + B(12)^1/2 = 1 here is the system

A(12)^1/2 + B(12)^1/2 = 0 multiply thru by (12)^1/2
___________________________
2B(12)^1/2 = 1 add equations

B = (1/2)(12)^-1/2 solve for B

A = -(1/2)(12)^-1/2 solve for A [A and B are opposites]


(-1/2)∫(12)^-1/2 (12)^-1/2
__________________ + (1/2) ∫ ____________
x + 12^1/2 x - 12^1/2

[make substitutions]

= (-1/2)(12)^-1/2 ln (x + 12^1/2) + (1/2)(12)^-1/2 ln (x - 12^1/2) + C

[integrate]


(9.) ∫(x - 1)/(x + 1) dx


= ∫(x + 1 - 2)/(x + 1) dx write -1 as 1 - 2

= ∫dx - 2∫[dx/(x + 1)] divide and separate like this

= x - 2 ln (x + 1) + C integrate

(10.) ∫ (2x + 3)/(x - 4) dx

= (2)∫(x + 1.5)/(x - 4) dx factor 2 out front

= 2∫(x - 4 + 5.5)/(x - 4) dx write 1.5 as -4 + 5.5

= 2∫dx + 11∫[dx/(x - 4)] divide and separate like this

= 2x + 11 ln (x - 4) + C integrate

(11.) ∫(x + 1)/(x - 1) dx

= ∫(x - 1 + 2)/(x - 1) dx write 1 as -1 + 2

= ∫(x - 1)dx/(x - 1) + 2∫[dx/(x - 1) separate like this

= ∫dx + 2∫[dx/(x - 1) cancel

= x + 2 ln (x - 1) + C integrate

(12.) ∫ (x² + 4x + 5)dx/(x - 2)

2 | 1 4 5 use synthetic division
__| 2 12
_________________
1 6 17

= ∫x dx + 6∫dx + 17∫dx/(x - 2) here is the new form

= (1/2)x² + 6x + 17 ln (x - 2) + C integrate

(13.) ∫ (x³ - x)dx/(2x + 6)

= (1/2)∫(x³ - x)/(x + 3) factor like this

1 0 -1 0 use synthetic division

-3| -3 9 -24
__| ____________________
1 -3 8 -24

= (1/2)∫(x² - 3x + 8)dx - 12∫[1/(x + 3)]dx

[here is the new form]

= (1/2)(1/3)x³ - (3/4)x² + 4x - 12 ln (x + 3) + C integrate

= (1/6)x³ - (3/4)x² + 4x - 12 ln (x + 3) + C multiply