If int(e^(4x)-1)/(e^(2x))log((e^(2x)+1)/...

If `int(e^(4x)-1)/(e^(2x))log((e^(2x)+1)/(e^(2x-1)))dx=(t^(2))/(2)logt-(t^(2))/(4)-(u^(2))/(2)logu+(u^(2))/(4)+C,` then

`u=e^(x)+e^(-x)`

`u=e^(x)-e^(-x)`

`t=e^(x)+e^(-x)`

`t=e^(x)-e^(-x)`

Text Solution

Verified by Experts

The correct Answer is:

B, C

`I=int{(e^(2x)-e^(-2x))ln(e^(x)+e^(-x))-(e^(2x)-e^(-2x))ln(e^(x)-e^(-x))}dx`
`=int tln t dt- int u ln u du ("where t"=e^(x)+e^(-x) and u=e^(x)-e^(-x))`
`=(t^(2))/(2)ln t-(t^(2))/(4)-(u^(2))/(2)ln u+(u^(2))/(4)+C`

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If int(e^(4x)-1)/(e^(2)x)log((e^(2x)+1)/(e^(2x)-1))dx (t^(2))/(2)logt-(t^(2))/(4)+(u^(2))/(4)+C them

`t=e^(-x)-e^(x)u=e^(x)+e^(-x)`

`t=e^(x)-e^(-x),u=e^(x)+e^(-x)`

`t=e^(x)+e^(-x),u=e^(x)-e^(-x)`

none of these

int(e^(2x)+1)/(e^(2x)-1)dx=

`log(e^(x)-e^(-x))+alpha`

`log(e^(x)+e^(-x))+beta`

`log(e^(2x)-1)+x+gamma`

`log(e^(2x)-1)+delta`