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Q9  Integrate the rational functions \frac{3x+ 5 }{x^3 - x^2 - x +1 }

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Given function \frac{3x+ 5 }{x^3 - x^2 - x +1 }

can be rewritten as \frac{3x+ 5 }{x^3 - x^2 - x +1 } = \frac{3x+5}{(x-1)^2(x+1)}

Partial function of this function:

\frac{3x+5}{(x-1)^2(x+1)}= \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}

3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2 

3x+5 = A(x^2-1)+B(x+1)+C(x^2+1-2x)      ................(1)

Now, putting x=1 in the above equation, we get

B =4

By equating the coefficients of x^2 and x , we get

A+C=0

B-2C =3

then after solving, we get

A= -\frac{1}{2}\ and\ C=\frac{1}{2}

Therefore, 

\frac{3x+5}{(x-1)^2(x+1)}= \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}

\int \frac{3x+5}{(x-1)^2(x+1)}dx= \frac{-1}{2}\int \frac{1}{(x-1)}dx+4\int \frac{1}{(x-1)^2} dx+\frac{1}{2}\int \frac{1}{(x+1)}dx

= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C

=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C

 

 

Posted by

Divya Prakash Singh

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