Integrate `(1-x^2)/(x(1-2x))`

Explanation

Here the degree of numerator is equal to the degree of denominatior . So we divide the numerator by the denominator.
`" Then " (1-x^(2))/(x(1-2x))=(x^(2)-1)/(2x^(2)-x)=(1)/(2) +((1)/(2)x-1)/(2x^(2)-1)`
` :. I= int(1)/(2) dx+int((1)/(2)x-1)/((2x^(2)-x))dx`
`rArr i=I_(1)+I_(2)`
`" Where "I_(1) = int(1)/(2) dx " and " I_(2) = int((1)/(2)x-1)/((2x^(2)-x))dx`
`" Let " ((1)/(2)x-1)/((2x^(2)-x))=(A)/(x)+(B)/((2x+1))` lt brgt `rArr ((1)/(2)x-1)/((2x^(2)-x))=(A(2x-1)+Bx)/(x(2x-1)`
`rArr (1)/(2)x-1 =A(2x-1) +Bx`
` x=0 " then " -1 =A(-1)+0 rArr A=1`
`x=(1)/(2) " then " (1)/(4) -1=0|B((1)/(2))rArr B=-(3)/(2)`
`:. I_(2)=int[(1)/(x)-(3)/(2(2x-1))]dx`
` = int (1)/(x) dx -(3)/(2) int(1)/(2x-1)dx`
`rArr I_(2) =log x-(3)/(2)(log |2x-1|)/(2)+c_(2)`
put the values of `I_(1)" and " I_(2)` in equation (1),
`I=(1)/(2)x+log x-(3)/(4) log |2x-1|+C`
`(C_(1)+C_(2)=C)`