partial integration help, please? if $\phi(x)={1\over x}\int\nolimits_0^x F(t)dt$ and $F(x):=\int_0^x f(t)dt$ ,how does $$\phi'(x)=-{1\over x^2}\int_0^xF(t)dt +{1\over x}F(x)={1\over x^2}\int_0^x t f(t)dt\ ?$$
Please explain step by step since I am confused how to get to the last two equalities. 
 A: Here are two hints.  Give it a try, I can expand on the details for Hint 2 if needed.
Hint 1:
$$
\phi'(x)=\frac{d}{dx} \frac{1}{x}
\int_0^x F(t)dt\\
=\left(\frac{d}{dx}\frac{1}{x}\right)
\int_0^xF(t)dt+\frac{1}{x}
\left(\frac{d}{dx}\int_0^x F(t)dt\right).
$$
Then the fundamental theorem of calculus says $$\left(\frac{d}{dx}\int_0^x F(t)dt\right)=F(x).$$  Also,  $$\frac{d}{dx}\frac{1}{x}=\frac{-1}{x^2}.$$
Hint 2:  For any nice function $f$, we have that $$\int_0^x \int_0^t f(u) du dt= \int_0^x (x-t)f(t)dt.$$  This is a case of Cauchy's formula for repeated integration.
Hope that helps,
A: For our purposes, we may assume that $f$ is continuous.
Step 1:
$$
\phi '(x) = \bigg(\frac{d}{{\,dx}}\frac{1}{x}\bigg)\int_0^x {F(t)\,dt}  + \frac{1}{x}\frac{d}{{\,dx}}\int_0^x {F(t)\,dt}. 
$$
Step 2:
$$
\phi '(x) = - \frac{1}{{x^2 }}\int_0^x {F(t)\,dt}  + \frac{1}{x}F(x).
$$
Step 3: 
$$
x^2 \phi '(x) =  - \int_0^x {F(t)dt}  + xF(x) = xF(x)  - \int_0^x {F(t)dt}.
$$
Step 4 -- integration by parts:
$$
\int_0^x {tf(t) \,dt}  = tF(t) \big|_0^x  - \int_0^x {\bigg(\frac{d}{{\,dt}}t \bigg)F(t)\,dt} = xF(x) - 0F(0) - \int_0^x {1F(t)\,dt}  = xF(x) - \int_0^x {F(t)\,dt}. 
$$
Step 5: It follows that
$$
x^2 \phi '(x) = \int_0^x {tf(t) \,dt}.
$$
Step 6: It follows that
$$
\phi '(x) = \frac{1}{{x^2 }}\int_0^x {tf(t) \,dt}.
$$
EDIT: For the integration by parts (beginning of Step 4), note that $F$ is an antiderivative of $f$, since, by the Fundamental theorem of calculus,
$$
F'(x) = \frac{d}{{dx}}F(x) = \frac{d}{{dx}}\int_0^x {f(t)dt}  = f(x).
$$
(Here we used the assumption that $f$ is continuous.)
