`int(2e^t)/(e^(3t)-6e^(2t)+11 e^t-6)dt`

The function f(x)=int_(-2015)^(x)t(e^(t)-e^(2))(e^(t)-1)(t+2014)^(2015)(t-2015)^(2016)(t-2016)^(2017) dt has

The locus of point ((e^(t)+e^(-t))/(2),(e^(t)-e^(-t))/(2)) is a hyperbola with eccentricity

if int(e^(t)dt)/(1+t)=a then int e^(t)(dt)/((1+t)^(2))=

A = [{:(e^(t), e^(-t)"cos"t, e^(-t)"sin"t),(e^(t)-e^(-t), "cos"t-e^(-t)"sin"t, -e^(-t)"sin"t + e^(-t)"cos"t),(e^(t), 2e^(-t)"sin"t, -2e^(-t)"cos"t):}]"then A is"

The equations x=(e^t+e^(-t))/2,y=(e^(t)-e^(-t))/2, t inR represent :

The locus of the point ( (e^(t) +e^(-t))/( 2),(e^t-e^(-t))/(2)) is a hyperbola of eccentricity

The locus of the point ( (e^(t) +e^(-t))/( 2),(e^t-e^(-t))/(2)) is a hyperbola of eccentricity

If int_(0)^(1)(e^(t))/(1+t)dt=a, then find the value of int_(0)^(1)(e^(t))/((1+t)^(2))dt in terms of a