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Inte= grals of Rational Functions
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In this section we will show how to integrate rational functions, that is, functions of the for= m          = ;       
P(x)
Q(x)=
where= P and Q are polynomials.
Inte= grals of Rational Functions
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In this section we will show how to integrate rational functions, that is, functions of the for= m          = ;       
P(x)
Q(x)=
where= P and Q are polynomials.
We use the decomposition theorem:
Given reduced P/Q where deg P < deg Q,
then P/Q =3D F1 + F= 2=  + .= . + Fn=  whe= re
  = ;
Fi =3D=         = ;       or 
A
 (ax + b)k=  
Ax += B
(ax<= /span>2 + bx + c)k=  
and that (ax + b)k or (ax2 + b= x + c)k=  are= factors in the factorization of Q(x).
Inte= grals of Rational Functions
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Therefore finding the integrals of rational functions are reduced to finding integrals of          = ;       
 or 
A
 (ax + b)k=  
Ax += B
(ax<= /span>2 + bx + c)k=  
Inte= grals of Rational Functions
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Therefore finding the integrals of rational functions are reduced to finding integrals of          = ;       
 or 
A
 (ax + b)k=  
Ax += B
(ax<= /span>2 + bx + c)k=  
Inte= grals of Rational Functions
The integrals ∫        = ;      are straight forward
with the substitution method by setting
u =3D ax + b.  
dx
 (ax + b)k=  
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Therefore finding the integrals of rational functions are reduced to finding integrals of          = ;       
 or 
A
 (ax + b)k=  
Ax += B
(ax<= /span>2 + bx + c)k=  
Inte= grals of Rational Functions
The integrals ∫        = ;      are straight forward
with the substitution method by setting
u =3D ax + b.  
dx
 (ax + b)k=  
To integrate ∫        = ;            dx, we need to
complete the square for the denominator. 
Ax += B
(ax<= /span>2 + bx + c)k=  
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x      ) + 10
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
Hence        
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
Set u =3D x + 3     
   &#= 13;
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
substitution           
Hence        
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
Set u =3D x + 3     
      x =3D u &= #8211; 3
    dx =3D du
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
substitution           
Hence        
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
Set u =3D x + 3     
      x =3D u &= #8211; 3
    dx =3D du
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
substitution           
=3D
 ∫        = ;    
u – 3
u2 + 1  
du
Hence        
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
Set u =3D x + 3     
      x =3D u &= #8211; 3
    dx =3D du
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
substitution           
=3D
 ∫        = ;    
u – 3
u2 + 1  
du
=3D
 ∫        = ;    
u 
u2 + 1  
du
 – 3
∫        = ;    
1 
u2 + 1  
du
Hence        
Inte= grals of Rational Functions
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x
x2 + 6x + 10  
Example:  Find ∫        = ;    
dx
x2 + 6x + 10 - its irreducible.=
Complete the square on x2 + 6x + 10 =3D  (x + 6x + 9) + 10 – 9
=3D  (x + 3)2 + 1
x
x2 + 6x + 10  
∫        = ;    
dx
Set u =3D x + 3     
      x =3D u &= #8211; 3
    dx =3D du
=3D
 ∫        = ;    
x
(x + 3)2 + 1  
dx
substitution           
=3D
 ∫        = ;    
u – 3
u2 + 1  
du
=3D
 ∫        = ;    
u 
u2 + 1  
du
 – 3
∫        = ;    
1 
u2 + 1  
du
Hence        
Inte= grals of Rational Functions
Set w =3D u2 + 1
 substitution           
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
=3D
½ ∫        = ;    
1 
w
 – 3 tan-1(u) + c =
dw
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
=3D
½ ∫        = ;    
1 
w
 – 3 tan-1(u) + c =
dw
=3D ½ Ln(w) – 3 tan-1(x + 3) + c
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
=3D
½ ∫        = ;    
1 
w
 – 3 tan-1(u) + c =
dw
=3D ½ Ln(w) – 3 tan-1(x + 3) + c
=3D ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Inte= grals of Rational Functions
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
=3D
½ ∫        = ;    
1 
w
 – 3 tan-1(u) + c =
dw
=3D ½ Ln(w) – 3 tan-1(x + 3) + c
=3D ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Inte= grals of Rational Functions
3x
x2 + 7x + 10  
Example:  Find ∫        = ;    
dx
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Set w =3D u2 + 1
     
        = ;  =3D 2u
substitution           
dw
du
du =3D
dw
2u
=3D
 ∫        = ;    
u 
w
 – 3
∫        = ;    
1 
u2 + 1  
du
dw
2u
=3D
½ ∫        = ;    
1 
w
 – 3 tan-1(u) + c =
dw
=3D ½ Ln(w) – 3 tan-1(x + 3) + c
=3D ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Inte= grals of Rational Functions
3x
x2 + 7x + 10  
Example:  Find ∫        = ;    
dx
Since x2 + 7x + 10 =3D (x + 2)(x + 5), we can = = decompose the ration= al expression.   
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