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int (x^(2)(1-"ln"x))/(("ln"^(4)x-x^(4))d...

`int (x^(2)(1-"ln"x))/(("ln"^(4)x-x^(4))dx` is equal to

A

`(1)/(2) " In" ((x)/("In "x))-(1)/(4) "In"("In"^(2)x-x^(2))+C`

B

`(1)/(4)"In"(("In"x-x)/("In "x+x))-(1)/(2)tan^(-1)(("In "x)/(x))+C`

C

`(1)/(4) "In"(("In "x+x)/("In "x-x))+(1)/(2)tan^(-1)(("In "x)/(x))+C`

D

`(1)/(4) "In"(("In "x-x)/("In "x+x))+(1)/(2)tan^(-1)(("In "x)/(x))+C`

Text Solution

Verified by Experts

The correct Answer is:
b

Let `I=int(x^(2)(1-"In "x))/(("In "x)^(4)-x^(4))dx` Then,
`I=int(x^(2)(1-"In "x))/((("In "x)^(2)+x^(2))("In "x+x)("In "x-x))dx`
`rArrI=int((1-"In "x)/(x^(2)))/({(("In "x)/(x))^(2)+1}(("In "x)/(x)+1)(("In "x)/(x)-1))dx`
`rArrI=int(1)/((t^(2)+1)(t+1)(t-1))dt` , where `t=("In "x)/(x)`
`rArrI=int(1)/((t^(2)+1)(t^(2)-1))dt`
`rArrI=(1)/(2)int((1)/(t^(2)-1)-(1)/(t^(2)+1))dt=(1)/(4)"In"|(t-1)/(t+1)|-(1)/(2)tan^(-1)t+C`
`rArrI=(1)/(4)"In"|(("In"x)/(x)-1)/(("In "x)/(x)+1)|-(1)/(2)tan^(-1)(("In "x)/(x))+C`
`rArrI=(1)/(4)"In"|("In "x-x)/("In "x+x)|-(1)/(2)tan^(-1)(("In "x)/(x))+C`
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