For all values of , `int_(1//e)^(tanx) (t)/(1+t^(2))dt+int_(1//e)^(tanx) (dt)/(t(t+t^(2)))` has the value

Options

A. 0
B. 1
C. e
D. none of these

Answer

B

Explanation

Consider the integral `underset(1//e)overset(tanx)int (dt)/(t(1+t^(2)))`
Putting `t=(1)/(u)`, we get
`underset(1//e)overset(tanx)int (dt)/(t(1+t^(2)))underset(e)overset(tanx)int(-1//u^(2)du)/(1//u(1+1//u^(2)))=-underset(e)overset(tanx)int (u du)/(1+u^(2))`
`rArr underset(1//e)overset(cotx)int (dt)/(t(1+t^(2)))underset(e)overset(tanx)int(tdt)/(1+t^(2))={underset(e)overset(tanx)int(tdt)/(1+t^(2))+underset(1//e)overset(tanx)int(tdt)/(1+t^(2))}`
`rArr underset(1//e)overset(cotx)int (dt)/(t(1+t^(2)))underset(e)overset(1//e)int(t)/(1+t^(2))dt- underset(1//e)overset(tanx)int(t dt)/(1+t^(2))`
`:. underset(1//e)overset(tanx)int (dt)/(1+t^(2))dtunderset(e)overset(1//e)int t(1+t^(2))dt`
`=- underset(e)overset(1//e)int (t)/(1+t^(2))dt=-(1)/(2)[ log(1+t^(2))]_(e)^(1//e)`
`=-(1)/(2)[ log (1+(1)/(e^(2)))-log(1+e^(2))]`
`=-(1)/(2)[ log (e^(2)+1)-2loge-log(1+e^(2))]=loge=1`