`1/(x+1)-1/x <= 1/(x-1)-1/(x-2)`

tan^(-1).(x(x+1)+1)/((x+1)-x) =tan^(-1) (x^(2) +x^+1) = R.H.S

Solve by factorization: (x+1)/(x-1)-(x-1)/(x+1)=5/6,\ \ x!=1,\ -1

Solve by factorization: (x+1)/(x-1)-(x-1)/(x+1)=(5)/(6),x!=1,-1

For x>0 the sum of the series (1)/(1+x)-(1-x)/((1+x)^(2))+((1-x)^(2))/((1+x)^(3))-...oo is equal to

If y=(1+(1)/(x))^(x), then (dy)/(dx)=(1+(1)/(x))^(x){log(1+(1)/(x))-(1)/(x+1)}(b)(1+(1)/(x))^(x)log(1+(1)/(x))(c)(x+(1)/(x))^(x){log(x+1)-(x)/(x+1)}(d)(x+(1)/(x))^(x){log(1+(1)/(x))+(1)/(x+1)}

If y=x+(1)/(x+(1)/(x+(1)/(x+(1)/(x+...)))), prove that (dy)/(dx)=(1)/(2(x)/(x+(1)/(x+(1)/(x+1))))

If x ne pm 1 , then (x+1)/(x-1)-(x-1)/(x+1)=

If y=(1+1/x)^x , then (dy)/(dx)= (a) (1+1/x)^x{log(1+1/x)-1/(x+1)} (b) (1+1/x)^xlog(1+1/x) (c) (x+1/x)^x{log(x+1)-x/(x+1)} (d) (x+1/x)^x{log(1+1/x)+1/(x+1)}