`int_0^1 log(sqrt(1+x)+sqrt(1-x))dx=
`
(A) `1/2(log2-pi/2+1)`
(B) `1/2(log2+pi/2+1)`
(C) `1/2(log2+pi/2-1)`
(D) none of these

log((sqrt2-1)/(sqrt2+1))

int x(ln(x+sqrt(1+x^2))/sqrt(1+x^2)dx

If int_(0)^(1) (log(1+x)/(1+x^(2))dx=

int_0^1sqrt((1-x)/(1+x))dx= (a) (pi)/2 (b) (pi)/2-1 (c) (pi)/2+1 (d) pi+1

T h ev a l u eofint_1^((1+sqrt(5))/2)(x^2+1)/(x^4-x^2+1)log(1+x-1/x)dxi s (a) pi/8(log)_e2 (b) pi/2(log)_e2 (c) -pi/2(log)_e2 (d) none of these

Prove that : int_(0)^(1) (log x)/(sqrt(1-x^(2)))dx=-(pi)/(2)log 2

int_0^pi dx/(1+10^(cosx))+int_(-1)^1 log((2-x)/(2+x))dx= (A) pi/2 (B) -pi (C) 0 (D) none of these

Area lying between the curves y=tanx, y=cotx and x-axis, x in [0, pi/2] is (A) 1/2log2 (B) log2 (C) 2log(1/sqrt(2)) (D) none of these

int(log_e(x+1)-log_ex)/(x(x+1))dx is equal to (A) -1/2[log(x+1)^2-1/2logx]^2+log_e(x+1)log_ex+C (B) -[(log_e(x+1)-log_ex]^2 (C) c-1/2(log(1+1/x))^2 (D) none of these

Solve log_(2)(2sqrt(17-2x))=1 - log_(1//2)(x-1) .