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Integration using Partial fraction

Type I

When the denominator is expressible as a product of linear factors

\displaystyle \int { \frac { x-1 }{ \left( x+1 \right) \left( x-2 \right) } } dx

Solution:

\displaystyle Let\quad \int { \frac { x-1 }{ \left( x+1 \right) \left( x-2 \right) } } dx=\frac { A }{ \left( x+1 \right) } +\frac { B }{ \left( x-2 \right) } \qquad .......(i)\\ \Rightarrow x-1=A\left( x-2 \right) +B\left( x+1 \right)

Putting x-2=0 or x=2 in (ii), we get 1=3B \displaystyle \Rightarrow B=\frac { 1 }{ 3 } 

Putting x+1 =0 or x=-1 in (ii), we get -2=-3A \displaystyle \Rightarrow A=\frac { 2 }{ 3 } 

Substituting the values of A and B in (i) , we get

\displaystyle \frac { x-1 }{ \left( x+1 \right) \left( x-2 \right) } =\frac { 2 }{ 3 } .\frac { 1 }{ x+1 } +\frac { 1 }{ 3 } .\frac { 1 }{ x-2 } \\ So,\quad \int { \frac { x-1 }{ \left( x+1 \right) \left( x-2 \right) } } dx=\frac { 2 }{ 3 } \int { \frac { 1 }{ x+1 } } dx+\frac { 1 }{ 3 } \int { \frac { 1 }{ x-2 } } dx\\ =\frac { 2 }{ 3 } \log { \left| x+1 \right| } +\frac { 1 }{ 3 } \log { \left| x-2 \right| } +c

 

Type II

When denominator contains some repeating linear factors

\displaystyle \int { \frac { 3x+1 }{ { \left( { x-2 } \right) }^{ 2 }\left( x+2 \right) } } dx

Solution:

\displaystyle \frac { 3x+1 }{ { \left( x-2 \right) }^{ 2 }\left( x+2 \right) } =\frac { A }{ \left( x-2 \right) } +\frac { B }{ { \left( x-2 \right) }^{ 2 } } +\frac { C }{ \left( x+2 \right) } \qquad ......\left( i \right) \\ \Rightarrow 3x+1=A\left( x-2 \right) \left( x+2 \right) +B\left( x+2 \right) +C{ \left( x-2 \right) }^{ 2 }\qquad ......\left( ii \right)  

Putting x-2=0 i.e. x=2 in (ii), we get 7=4 B \displaystyle \Rightarrow B=\frac { 7 }{ 4 } 

Putting x+2 =0 i.e. x=-2 in (ii) we get -5=16 C \displaystyle \Rightarrow C=-\frac { 5 }{ 16 }  

Comparing coefficients of \displaystyle { x }^{ 2 }   on both sides of the identity (ii) we get

\displaystyle A+C=0\Rightarrow A=-C\Rightarrow A=\frac { 5 }{ 16 }  

Substituting the values of A,B and C in (i) we get

\displaystyle \frac { 3x+1 }{ { \left( x-2 \right) }^{ 2 }\left( x+2 \right) } =\frac { 5 }{ 16 } .\frac { 1 }{ \left( x-2 \right) } +\frac { 7 }{ 4 } \frac { 1 }{ { \left( x-2 \right) }^{ 2 } } -\frac { 5 }{ 16\left( x+2 \right) } \\ So,I=\int { \frac { 3x+1 }{ { \left( x-2 \right) }^{ 2 }\left( x+2 \right) } } dx=\frac { 5 }{ 16 } \int { \frac { 1 }{ \left( x-2 \right) } } dx+\frac { 7 }{ 4 } \int { \frac { 1 }{ { \left( x-2 \right) }^{ 2 } } } dx\frac { 5 }{ 16 } \int { \frac { 1 }{ \left( x+2 \right) } } dx\\ =\frac { 5 }{ 16 } \log { \left| x-2 \right| } -\frac { 7 }{ 4\left( x-2 \right) } -\frac { 5 }{ 16 } \log { \left| x+2 \right| } +c

 

Type III

The denominator contains irreducible quadratic factors

\displaystyle \int { \frac { x }{ \left( x+2 \right) \left( { x }^{ 2 }+4 \right) } } 

Solution:

\displaystyle Let\quad I=\int { \frac { x }{ \left( x-1 \right) \left( { x }^{ 2 }+4 \right) } } =\frac { A }{ x-1 } +\frac { Bx+C }{ { x }^{ 2 }+4 } \qquad ......\left( i \right) \\ \Rightarrow x=A\left( { x }^{ 2 }+4 \right) +\left( Bx+C \right) \left( x-1 \right) \qquad ......\left( ii \right) 

Putting x-1 in (ii), we get 1=5 A  Putting x=0 in (ii), we get 0= 4 A – C

Putting x =-1 in (ii), we get -1 = 5 A + 2 B -2 C

Substituting the values of A , B and C in (i), we obtain

\displaystyle \frac { x }{ \left( x-1 \right) \left( { x }^{ 2 }+4 \right) } =\frac { 1 }{ 5\left( x-1 \right) } +\frac { -\frac { 1 }{ 5 } x+\frac { 4 }{ 5 } }{ { x }^{ 2 }+4 }  

\displaystyle \Rightarrow \frac { x }{ \left( x-1 \right) \left( { x }^{ 2 }+4 \right) } =\frac { 1 }{ 5\left( x-1 \right) } -\frac { 1 }{ 5 } \frac { \left( x-4 \right) }{ \left( { x }^{ 2 }+4 \right) }  

\displaystyle \int { \frac { x }{ \left( x-1 \right) \left( { x }^{ 2 }+4 \right) } dx } =\frac { 1 }{ 5 } \int { \frac { 1 }{ \left( x-1 \right) } } dx-\frac { 1 }{ 5 } \int { \frac { \left( x-4 \right) }{ \left( { x }^{ 2 }+4 \right) } } dx\\ =\frac { 1 }{ 5 } \int { \frac { 1 }{ \left( x-1 \right) } } dx-\frac { 1 }{ 10 } \int { \frac { 2x }{ \left( { x }^{ 2 }+4 \right) } } dx+\frac { 4 }{ 5 } \int { \frac { 1 }{ \left( { x }^{ 2 }+4 \right) } } dx

\displaystyle =\frac { 1 }{ 5 } \log { \left| x-1 \right| } -\frac { 1 }{ 10 } \log { \left| { x }^{ 2 }+4 \right| } +\frac { 4 }{ 5 } \times \frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { x }{ 2 } \right) } +C\\ =\frac { 1 }{ 5 } \log { \left| x-1 \right| } -\frac { 1 }{ 10 } \log { \left| { x }^{ 2 }+4 \right| } +\frac { 2 }{ 5 } \tan ^{ -1 }{ \left( \frac { x }{ 2 } \right) } +C

 

Type IV

If rational function contains only even powers of x in both the numerator and denominator, then to resolve it into partial functions, we proceed as follows:

Step I  Put \displaystyle { x }^{ 2 }=y  in the given rational function

Step II Resolve the rational function obtained in step I into partial fractions.

Step III Replace y by \displaystyle { x }^{ 2 }

Example

\displaystyle \int { \frac { { x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) } } dx

Solution:

\displaystyle Let\quad { x }^{ 2 }=y\quad Then,\quad \frac { { x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) } \quad =\quad \frac { y }{ \left( y+1 \right) \left( y+4 \right) }

\displaystyle Let\quad \frac { y }{ \left( y+1 \right) \left( y+4 \right) } =\frac { 1 }{ 3\left( y+1 \right) } +\frac { 4 }{ 3\left( y+4 \right) } \qquad ....\left( i \right) \\ \Rightarrow y=A\left( y+4 \right) +B\left( y+1 \right) \qquad ......\left( ii \right)  

Putting y=-1 and y=-4 successively in (ii), we get \displaystyle A=-\frac { 1 }{ 3 } \quad and\quad B=\frac { 4 }{ 3 }  

Substituting the values of A and B in (i), we obtain

\displaystyle \frac { y }{ \left( y+1 \right) \left( y+4 \right) } =-\frac { 1 }{ 3\left( y+1 \right) } +\frac { 4 }{ 3\left( y+4 \right) } 

Replacing y by \displaystyle { x }^{ 2 }  , we obtain

\displaystyle \frac { { x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) } =-\frac { 1 }{ 3\left( { x }^{ 2 }+1 \right) } +\frac { 4 }{ 3\left( { x }^{ 2 }+4 \right) } \\ \therefore \int { \frac { { x }^{ 2 } }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right) } } dx=-\frac { 1 }{ 3 } \int { \frac { 1 }{ \left( { x }^{ 2 }+1 \right) } } dx+\frac { 4 }{ 3 } \int { \frac { 1 }{ \left( { x }^{ 2 }+4 \right) } } dx

\displaystyle =-\frac { 1 }{ 3 } \tan ^{ -1 }{ x } +\frac { 4 }{ 3 } \times \frac { 1 }{ 2 } \tan ^{ -1 }{ \left( \frac { x }{ 2 } \right) } +C\\ =-\frac { 1 }{ 3 } \tan ^{ -1 }{ x } +\frac { 2 }{ 3 } \tan ^{ -1 }{ \left( \frac { x }{ 2 } \right) } +C

 

 

April 18, 2019
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