How to compute this double integral with $dt$ and $dx$? How to calculate the following integral? Any help is appreciated because I am not sure if there are some theorems I haven't learned to solve the question. Thanks.
$\int_0^1 \frac{f(x)}{\sqrt{x}} dx $
where $f(x) = \int_1^x \frac{\ln(1+t)}{t} dt$

I just found the solution of the first two steps and it's as the following
And it is more confusing to me now.
$\int_0^1 \frac{1}{\sqrt{x}} \int_1^x \frac{\ln(1+t)}{t} dt dx  $
$= -\int_0^1  \frac{\ln(1+t)}{t} dt \int_t^0  \frac{1}{\sqrt{x}} dx $
$= -2\int_0^1  \frac{\ln(1+t)}{\sqrt{t}} dt $
... (From this step on, I guess I could solve it by letting $t=u^2$)
 A: $$\int_{0}^{1}\int_{1}^{x}\frac{\ln(1+t)}{t\sqrt{x}}\,dt\,dx=-\int_{0}^{1}\int_{x}^{1}\frac{\ln(1+t)}{t\sqrt{x}}\,dt\,dx$$.
Now you have to change the order of integration.
If you look at the $x-t$ plane then we are integrating over the triangle
Then the limits become:-
$$-\int_{0}^{1}\int_{0}^{t}\frac{\ln(1+t)}{t\sqrt{x}}\,dx\,dt$$.
Now you can integrate with $x$ first keeping $t$ as a constant
So the integral becomes :-
$$-2\int_{0}^{1}\frac{\ln(1+t)}{\sqrt{t}}dt$$.
Now we would like to use the Maclauirin expansion of $\ln(1+x)$ in $0<|x|<1$.
$\ln(1+x)=\sum_{r=1}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}$.
So we get $$-4\sum_{r=1}^{\infty}\int_{0}^{1}(-1)^{r-1}\frac{t^{r-\frac{1}{2}}}{r}\,dt=-4\sum_{r=1}^{\infty}\frac{(-1)^{r-1}}{r(r+\frac{1}{2})}$$
$$=8\sum_{r=1}^{\infty}(-1)^{r}\frac{1}{2r(2r+1)}=8\sum_{r=1}^{\infty}\frac{(-1)^{r}}{2r}-8\sum_{r=1}^{\infty}\frac{(-1)^{r}}{2r+1}$$.
The above step was possible as both sums are convergent by Leibnitz Test.
The 2nd sum is nothing but $$8\sum_{r=1}^{\infty}\frac{(-1)^{r}}{2r+1}=8\left(-\frac{1}{3}+\frac{1}{5}-.....\right)$$.
If you are familiar with the maclaurin expansion of $\arctan(x)$ . Then you will see that it is nothing but $8(\arctan(1)-1)=2\pi-8$ . (This is also a well known series represesntation of $\pi$.
The 1st sum is just $4\sum_{r=1}^{\infty}\frac{(-1)^{r}}{r}$. This is nothing but the Maclaurin expansion of $-2\ln(1+x)$ evaluated at $x=1$. So it is $-4\ln(2)$.
Hence we have our answer as :-
$$8-2\pi-4\ln(2)$$.
Edit:-
A simple IBP would work totally fine. It was silly of me to invoke such complicated techniques.
if you make the sub $t=u^{2}$ then you have $-4\int_{0}^{1}\ln(1+u^{2})\,du$.
Now apply IPB to easily solve this.
A: $$2\int_{0}^{1}\frac{f(x)}{2\sqrt{x}}dx$$
By parts leads to,
$$-2\int_{0}^{1}f'(x)\sqrt{x}dx$$
Since $f(x)=\int_{1}^{x}\frac{\ln(1+t)}{t}dt$, by the fundamental theorem of calculus we have that,
$$f'(x)=\frac{\ln(1+x)}{x}$$
Putting that expression inside our integral we have that,
$$-2\int_{0}^{1}\frac{\ln(1+t)}{\sqrt{x}}dx$$
By the series expansion of $\ln(1+x)$ we have,
$$-2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k}\int_{0}^{1}x^{k-\frac{1}{2}}dx$$
$$-2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k(k+0.5)}$$
$$-4\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k(2k+1)}$$
By partial fraction decomposition,
$$-4\left(\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k} -2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{2k+1}\right)$$
The both series can be evaluated,
First series can be solved with Dirichlet eta function,
$$\eta(s)=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n^{s}}$$
It converges for all $\Re(s)>0$.
Second series can be computed by subtracting all the even terms from the Dirichlet eta function evaluated at $1$.
Sum of all even terms is half the total sum.
Now you compute it.
