(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 + 121/2           x - 121/2     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 + 121/2                      x - 121/2

[make substitutions]

=  (-1/2)(12)-1/2 ln (x + 121/2) + (1/2)(12)-1/2 ln (x - 121/2) +  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 + 6dx + 17dx/(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