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Q 19  Integrate the rational functions \frac{2x }{( x^2 +1)( x^2 +3)}

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

Taking x^2 = t \Rightarrow 2xdx=dt

\therefore \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \frac{dt}{(t+1)(t+3)}               

The partial fraction of above equation,

\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}

1= A(t+3)+B(t+1)                                            ..............(1)                            

Now, substituting t = -3\ and\ t = -1 in equation (1), we get

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

\therefore\frac{1}{(t+3)(t+3)} = \frac{1}{2(t+1)}-\frac{1}{2(t+3)}

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

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

= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C

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

Posted by

Divya Prakash Singh

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