Integral Confusion (Review) How do I solve $$\int_{0}^{x}\ln(x+t)f(t)dt = -x^2\left[ 1+3\ln(x)\right ]$$ It's been awhile since I've done Calculus and I am currently reviewing. Though I am not sure what to do with this problem.
 A: Let us denote the integral in the LHS of the equation as follows:-
$$I_{f(t)}=\int_0^x\ln(x+t)f(t)dt$$
Examining the RHS of the equation in the question, note that there is a quadratic term in $x$ multiplied by a log term. 
As cnick has helpfully mentioned, we can use integration by parts, where we have a log term multiplied by another function. The log term is usually differentiated, so that the other term is integrated. 
From the RHS, the quadratic term in $x$ results from integrating a linear term in $x$, so let us set $g(t)=(x+t)$ and evaluate the following integral by parts (setting $u=\ln(x+t)$, $dv=g(t)=(x+t)$):-
$$I_{g(t)}=\int_0^x\ln(x+t)g(t)dt=\int_0^x(x+t)\ln(x+t)dt\\=\left[\frac{(x+t)^2}{2}\ln(x+t)\right]_0^x-\int_0^x\frac{(x+t)^2}{2}\frac{1}{(x+t)}dt\\=2x^2\ln(2x)-\frac{x^2}{2}\ln(x)-\left[\frac{(x+t)^2}{4}\right]_0^x\\=\frac{3x^2}{2}\ln(x)+2x^2\ln2-\frac{3x^2}{4}$$
The task is now to transform the above expression to match the RHS of the original equation as follows:-
$$-2I_{g(t)}=I_{-2g(t)}=-3x^2\ln(x)-4x^2\ln2+\frac{3x^2}{2}=-3x^2\ln x+\left(\frac{3}{2}-4\ln2\right)x^2 \\\Rightarrow I_{-2g(t)}+\color{blue}{\left(4\ln2-\frac{5}{2}\right)x^2}=-x^2[1+3\ln x]$$
The term highlighted in blue can be expressed as the following definite integral:-
$$\left(4\ln2-\frac{5}{2}\right)x^2=\left(4\ln2-\frac{5}{2}\right)\int_0^x (2t)dt\\=\left(4\ln2-\frac{5}{2}\right)\int_0^x\ln(x+t)\frac{2t}{\ln(x+t)}dt$$
Thus combining the term $\left(4\ln2-\frac{5}{2}\right)\frac{2t}{\ln(x+t)}$ in the second integral with $-2g(t)$ we obtain a solution for $f(t)$ as below:-
$$f(t)=-2g(t)+\left(4\ln2-\frac{5}{2}\right)\frac{2t}{\ln(x+t)}\\\Rightarrow f(t)=-2(x+t)+\left(4\ln2-\frac{5}{2}\right)\frac{2t}{\ln(x+t)}$$ 
A: There is not a unique solution for this interesting problem. The question asks to solve for $f(t)$, considering $x$ as a constant. First note that, to generate a second-degree integral, $f(t)$ has to be a first-degree function. Setting  $f(t)=t$, the indefinite integral is
$$\frac 14 [(2xt-t^2)-2(x^2-t^2)\ln(x+t)]+c$$
and the definite integral calculated between $0$ and $x$ is
$$\frac 14 x^2 [ 1+2 \ln(x) ]$$
Simply multiplying by $-4$ (i.e., setting $f(t)=-4t$) yields a definite integral of $- x^2 [ 1+2 \ln(x)]$.
Considering that the difference between this result and that stated in the question is $-x^2\ln(x)$, we have to find a function $g(t)$ so that setting $f(t)=-4t +g(t)$ the integral gives the correct result. This means that
$$\displaystyle\int_0^x \ln(x+t)g(t)dt=-x^2\ln(x)$$
We can now look for a function $h(t)$ (resulting from calculation of an indefinite integral) so that $h(x)-h(0)= -x^2\ln(x)$. Since there is not a unique function that satisfies this condition, we can choose a simple one. For example, considering $\ln(x)$ as a constant, we can set $h(t)=-t^2\ln(x)$. Its derivative is $-2t\ln(x)$. We can then write 
$$\ln(x+t)g(t) =-2t\ln(x)$$
which leads to 
$$g(t)=-2t\ln(x)/\ln(x+t)$$
The final result is then
$$f(t)=-4t -2t\ln(x)/\ln(x+t)$$
As a confirmation, inputing  $\ln(x+t) [-4t -2t\ln(x)/\ln(x+t)]$ in Wolfram, using the appropriate letters, the resulting integral is $2x^2\ln(x+t)-t^2\ln(x) -2t^2\ln(x+t) -2xt+t^2$. Calculating the definite integral between $0$ and $x$, we get $-x^2[ 1+3\ln(x) ]$.
As already mentioned, this is not the only solution. For instance, we could set $h(t)=(x^2-t^2)\ln(x)$, which satisfies the condition $h(x)-h(0)= -x^2\ln(x)$ as well. Proceding as shown above, we could obtain a different valid solution.
