Evaluate : (i) `int(dx)/((1-4x^(2)))`
(ii) `int(dx)/((32-2x^(2)))`
(iii) `int(x^(2))/((1-x^(6)))dx`

Explanation

We have (i) `int(dx)/((1-4x^(2)))=(1)/(4).int(dx)/(((1)/(4)-x^(2)))`
`=(1)/(4).int(dx)/({((1)/(2))^(2)-x^(2)})`
`=(1)/(4)*(1)/((2xx(1)/(2)))*log|((1)/(2)+x)/((1)/(2)-x)|+C [becauseint(dx)/((a^(2)-x^(2)))=(1)/(2a)log|(a+x)/(a-x)|+C]`
`=(1)/(4)log|(1+2x)/(1-2x)|+C`
(ii) `int(dx)/((32-2x^(2)))=(1)/(2)*int(dx)/((16-x^(2)))` `=(1)/(2)*int(dx)/({(4)^(2)-x^(2)})`
`=(1)/(2)*(1)/((2xx4))log|(4+x)/(4-x)|+C`
`[becauseint(dx)/((a^(2)-x^(2)))=(1)/(2a)log|(a+x)/(a-x)|+C]`
`=(1)/(16)log|(4+x)/(4-x)|+C`.
(iii) Putting `x^(3)=t"and "3x^(2)dx=dt`, we get
`int(x^(2))/((1-x^(6)))dx=(1)/(3)*int(1)/((1-t^(2)))dt`
`=(1)/(3)*(1)/((2xx1))*log|(1+t)/(1-t)|+C`
`[becauseint(dx)/((a^(2)-x^(2)))=(1)/(2a)log|(a+x)/(a-x)|+C]`
`=(1)/(6)log|(1+x^(3))/(1-x^(3))|+C`.