Let [tex]c\in(0,\infty)[/tex]. Then the definite integral can be split up at [tex]t=c[/tex] so that
[tex]\displaystyle F(x)=\int_{t=2x}^{t=5x}\frac{\mathrm dt}t[/tex]
[tex]\displaystyle=\int_{t=2x}^{t=c}\frac{\mathrm dt}t+\int_{t=c}^{t=5x}\frac{\mathrm dt}t[/tex]
[tex]\displaystyle=-\int_{t=c}^{t=2x}\frac{\mathrm dt}t+\int_{t=c}^{t=5x}\frac{\mathrm dt}t[/tex]
Now take the derivative. By the fundamental theorem of calculus, you have
[tex]\displaystyle\frac{\mathrm dF}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\left[\int_{t=c}^{t=5x}\frac{\mathrm dt}t-\int_{t=c}^{t=2x}\frac{\mathrm dt}t\right][/tex]
[tex]=\displaystyle\frac1{5x}\dfrac{\mathrm d}{\mathrm dx}[5x]-\frac1{2x}\dfrac{\mathrm d}{\mathrm dx}[2x][/tex]
[tex]=\dfrac5{5x}-\dfrac2{2x}[/tex]
[tex]=\dfrac1x-\dfrac1x[/tex]
[tex]=0[/tex]
Then integrating with respect to [tex]x[/tex], we recover [tex]F(x)[/tex] and find that
[tex]\displaystyle\int\dfrac{\mathrm dF}{\mathrm dx}\,\mathrm dx=\int0\,\mathrm dx[/tex]
[tex]F(x)=C[/tex]
where [tex]C[/tex] is an arbitrary constant.
