[tex] \rm \int_{4}^8 \frac{f'(x)}{[f(x)] {}^2 } dx \\ [/tex]
put f(x)=u
then f'(x)dx=du
[tex] \rm \int_{f(4)}^{f(8)} \frac{du}{u {}^2 } = \frac{ - 1}{u} \bigg|_{ \frac{1}4 }^{ \frac{1}8 } \\ [/tex]
[tex]=-2+4=2[/tex]
Consider
[tex] \rm \int_{4}^8 \bigg(\frac{f'(x)}{[f(x)] {}^2 } + k \bigg)^{2} dx\\[/tex]
[tex] = \rm \int_{4}^8 \frac{(f'(x) {)}^{2} }{(f(x)){}^4} dx + 2k \int_{4}^8 \frac{f'(x) }{(f(x)){}^2} + {k}^{2}\int_{4}^8 dx\\[/tex]
[tex] = \rm1+2k \cdot2 + {k}^{2} \cdot4[/tex]
[tex] = \rm1 + 4k + 4 {k}^{2} [/tex]
[tex] = \rm(2k + 1 {)}^{2} [/tex]
[tex]\rm \int_{4}^8 \bigg(\frac{f'(x)}{[f(x)] {}^2 } + k \bigg)^{2} dx = (2k + {1)}^{2} \\ [/tex]
put k=-1/2
[tex]\rm \int_{4}^8 \bigg(\frac{f'(x)}{[f(x)] {}^2 } - \frac{1}{2} \bigg)^{2} dx = 0 \\ [/tex]
[tex]\rm \frac{f'(x)}{[f(x)] {}^2 } = \frac{1}{2} \\ [/tex]
[tex]\rm \int\frac{f'(x)}{[f(x)] {}^2 } dx = \int \frac{1}2dx \\ [/tex]
[tex] \rm \frac{ - 1}{f(x)} = \frac{x}2 + c\\ [/tex]
put x=4
[tex] \rm \frac{ - 1}{f(4)} = 2 + c \\ [/tex]
[tex] \rm - 4 = 2 + c [/tex]
[tex] \rm c=-6[/tex]
[tex] \rm \frac{ - 1}{f(x)} = \frac{x}{2} - 6\\ [/tex]
[tex] \rm f(x) = \frac{2}{12 - x} \\ [/tex]
put x=6
[tex] \rm f(6) = \frac{2}{12 - 6} \\ [/tex]
[tex]\rm f(6) = \frac{1}{3} \\ [/tex]
