" *49."(1-x^(2))/(x(1-2x))

tan((1)/(2) sin ^(-1)""(2x)/(1+x^(2))+(1)/(2)cos^(-1)((1-x^(2))/(1+x^(2))))=(2x)/(1-x^(2))(|x|ne 1)

If 49x^(2)-b=(7x+(1)/(2))(7x-(1)/(2)), then the value of b is

If 49x^(2)-b=(7x+(1)/(2))(7x-(1)/(2)), then the value of b is

(0.5x+49)/(10x^(2)+x-2)<(1)/(2)

If e_(1) is the eccentricity of the hyperbola (x^(2))/(36)-(y^2)/(49) =1 and e_(2) is the eccentricity of the hyperbola (x^(2))/(36)-(y^2)/(49) =-1 then

int (x^(49)Tan^(-1) (x^(50)))/((1+x^(100)))dx=k (Tan^(-1)(x^(50))^(2)+c rArr k=

int(x^(49)tan^(-1)(x^(50)))/((1+x^(100)))dx=k[tan^(-1)(x^(50))]^(2)+C, , then k is equal to

int_( then k)^( if )(x^(49)tan^(-1)(x^(50)))/(1+x^(100))dx=k*[tan^(-1)(x^(50))]^(2)+c