>0. This inequality holds when ⇒ Both x−1>0 and x+1>0⇒x>1 ⇒ Both x−1<0 and x+1<0⇒x<−1 So, the domain of f(x) is (−∞,−1)∪(1,∞) Now, f′(x)=‌
d
dx
[x+log(‌
x−1
x+1
)] ‌⇒1+‌
(x+1)
(x−1)
⋅‌
(x+1)⋅1−1⋅(x−1)
(x+1)2
‌⇒1+‌
(x+1)−(x−1)
(x−1)(x+1)
‌⇒1+‌
2
(x−1)(x+1)
Since, x∈(−∞,−1)∪(1,∞), we have x2>1 ⇒x2−1>0 Thus, ‌
2
x2−1
>0 So, f′(x)=1+‌
2
x2−1
>1+0=1 Since, f′(x)>0 for all x in the domain, the function f(x) is monotonically increasing.