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Q18  Integrate the rational functions \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

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Given function \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

We can rewrite it as: \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}

Partial fraction of above equation,

\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}

4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)

4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D

4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)                                  

Now, equating the coefficients of x^3, x^2, x and constant term, we get

A+C=0, B+D = 4, 4A+3C = 0, 4B+3D =10

After solving these equations, we get

A= 0, B =-2, C=0,\and\ D=6

\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}

\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )

\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx

= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx

= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C

= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C

 

Posted by

Divya Prakash Singh

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