Antiderivative of lebesgue integrable function

Let $f:[0,b]\to \mathbb{R}$ be Lebesgue integrable. We define

$g(x)={\int}_{x}^{b}\frac{f(t)}{t}dt,\phantom{\rule{1em}{0ex}}0<x\le b.$

Show that g(x) is Lebesgue integrable in [0,b]. Also show that

${\int}_{0}^{b}g(x)dx={\int}_{0}^{b}f(t)dt.$

If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that,

${\int}_{0}^{b}{\textstyle [}{g}^{\prime}(x)-\frac{f(x)}{x}{\textstyle ]}dx=0$, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.

Let $f:[0,b]\to \mathbb{R}$ be Lebesgue integrable. We define

$g(x)={\int}_{x}^{b}\frac{f(t)}{t}dt,\phantom{\rule{1em}{0ex}}0<x\le b.$

Show that g(x) is Lebesgue integrable in [0,b]. Also show that

${\int}_{0}^{b}g(x)dx={\int}_{0}^{b}f(t)dt.$

If we show the asked equality then it is obvious that g is Lesbegue integrable, since f is. Now, I have try to show the equality using the fact that,

${\int}_{0}^{b}{\textstyle [}{g}^{\prime}(x)-\frac{f(x)}{x}{\textstyle ]}dx=0$, since g is the antiderivative of f but I had no success. I am pleased to know other ideas to approach this problem.