f(x)+c Let u=x50, then du=50x49‌dx ‌so,I=∫x49[tan−1x50+‌
x50
1+x100
]‌dx ‌=∫x49[tan−1(u)+‌
u
1+u2
]⋅‌
du
50x49
‌=‌
1
50
‌∫(tan−1(u)+‌
u
1+u2
)du ‌=‌
1
50
[∫tan−1(u)du+∫‌
u
1+u2
du] ‌=‌
1
50
[utan−1(u)−∫‌
u
1+u2
du+∫‌
u
1+u2
du] [using product rule] ‌=‌
1
50
[utan−1(u)]+c ‌=‌
1
50
⋅x50⋅tan−1(x50)+c Comparing to original equation, we get n=50,k=50,f(x)=tan−1(x50) Now, f(x)−f(‌k√xn) ‌=f(x)−f(xn∕k) ‌=tan−1(x50)−f(x50∕50) ‌=tan−1(x50)−f(x) ‌=tan−1(x50)−tan−1(x50)=0 ∴f(x)−f(‌k√xn)=0 Now, k−n=50−50=0 From Eqs. (i) and (ii), we get f(x)−f(‌k√xn)=k−n
