IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

iitkgp_bipin

F(x) = _[1]

^[x] {logt / (1 + t)}.dt + _[1]

^[1/x] {log(t) / (1 + t)}.dt

Differentiating it wrt x :

F'(x) = {logx/(1+x)}.dx/dx + {log(1/x) / (1 + 1/x)}.d(1/x)/dx

F'(x) = {logx/(1+x)} + {logx/(1+x).x}

F'(x) = logx/x

F(x) =

(logx/x).dx

Differentiating it by parts we get :

F(x) = (logx)² -

(logx/x).dx + c = (logx)² - F(x) + c

2F(x) = (logx)² + c

From the original equation at x=1, F(x) = 0.

so, 0 = (log1)² + c

c = 0

so F(x) = (logx)²/2

F(e) = (loge)²/2 = 1/2

Ask & Discuss Questions with Community & Experts