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Q6  Integrate the rational functions \frac{1- x^2 }{ x ( 1- 2x )}

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

Integral is not a proper fraction so,

Therefore, on dividing (1-x^2) by x(1-2x), we get

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

Partial function of this function:

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

(2-x) =A(1-2x)+Bx                                  ...........(1)

Now, substituting x=0\ and\ \frac{1}{2}  respectively in equation (1), we get

A =2,\ B=3

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

No, substituting in equation (1) we get

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

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

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

=\frac{x}{2}+\log|x| -\frac{3}{4}\log|1-2x| +C

Posted by

Divya Prakash Singh

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