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Q12  Integrate the rational functions \frac{x^3 + x +1}{ x^2-1}

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

As the given integral is not a proper fraction.

So, we divide (x^3+x+1) by x^2-1, we get

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

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

2x+1 ={A}{(x-1)} +{B}{(x+1)}                           ....................(1)

Now, substituting x =1\ and\ x=-1 in equation (1), we get

A =\frac{1}{2}\ and\ B=\frac{3}{2}

Therefore, 

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

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

= \frac{x^2}{2}+\frac{1}{2}\log|x+1| +\frac{3}{2}\log|x-1| +C

 

Posted by

Divya Prakash Singh

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