Antiderivative of $\log(x)$ without Parts I understand how the antiderivative  of $\log(x)$ can be obtained by Integration by Parts (i.e. product rule), but I was wondering how-if at all- it could be obtained only using sum/difference rule and substitution/chain rule.
 A: $$\int_1^t \ln(x)\,dx = \int_1^t\int_1^x\frac 1u\,du\,dx = \int_1^t\int_u^t\frac 1u\,dx\,du = \int_1^t\frac{t-u}{u}\,du = {\large[}t\ln(u)-u{\large]}_{1}^t$$
A: There's a fun formula relating the integral of an invertible function and the integral of its inverse, namely
$$
bf(b)-af(a)=\int_a^bf(t)dt+\int_{f(a)}^{f(b)}f^{-1}(t)dt
$$
Which is easily seen from a picture. Then setting $b=x$, $0<x<a$ and $f(t)=\log t$ gives
$$
x\log x-a\log a=\int_a^x\log tdt+\int_{\log a}^{\log x}e^tdt\\
\Rightarrow x\log x-a\log a-\int_{\log a}^{\log x}e^tdt=\int_a^x\log tdt\\
\Rightarrow x\log x-x+a-a\log a=\int_a^x\log tdt\\
\Rightarrow x\log x-x+c=\int_a^x\log tdt
$$
A: We have:
$$\int_0^{t} x^a dx = \frac{t^{a+1}}{a+1}$$
Differentiating w.r.t. $a$ gives:
$$\int_0^{t} x^a \log(x)dx = \frac{t^{a+1}\left[(a+1)\log(t)-1\right]}{(a+1)^2}$$
The result then follows if we put $a = 0$.
A: Fermat's Method of Exhaustion can be used to obliterate this problem from first principles.
However, we will require the evaluation of a limit, which will be done so in advance.
Preliminary:
$\displaystyle \lim_{r \to 1} \frac{r \log{r}}{1-r} = -1$
Proof:
Consider the elementary inequality which holds true for all $r>0$:
$\displaystyle r-1 \le \log{r} \le 1-\frac{1}{r}$
The result follows by the squeeze theorem.
Consider a real number $0<r<1$, which will be used to construct the infinite geometric sequence of intervals.
Construct the lines $x = b, x = br, x = br^2, x = br^3, \cdots$ all the way to infinity.
The region bounded by $\log x, x = br^n, x = br^{n+1}$ is bounded strictly from below and above by the following inequality:
$\displaystyle b(1-r)r^{n+1} (\log{b}+(n+1)\log{r}) < \int_{br^{n+1}}^{br^n} \log{x}\, \text{d}x < b(1-r)r^n (\log{b}+n\log{r})$
Summing up these inequalities from $n=0$ to $\infty$ will give us strict upper and lower bounds on the integral.
After some elementary Arithmetico-Geometric Summation, we have the following:
$\displaystyle r \, b \, \log{b} + b\,\frac{r \log{r}}{1-r} < \int_0^b \log{x} \, \text{d}x < b\log{b} + b\,\frac{r \log{r}}{1-r}$
Take the limit as $r$ approaches $1$, and by the squeeze theorem, it can then be concluded:
$\displaystyle \int_0^b \log{x} \, \text{d}x = b\log{b} - b$
A: This may not necessarily be what you're looking for, but I got a kick out of it so I figured I'd share. We may evaluate the antiderivative by using integration by parts indirectly.
We seek the integral 
$$I_f(x)=\int f^{-1}(x)dx.$$
We set $x=f(u)$ to get
$$I_f(x)=\int uf'(u)du.$$
Integration by parts gives
$$I_f(x)=uf(u)-\int f(u)du$$
which is
$$I_f(x)=xf^{-1}(x)-F\circ f^{-1}(x)+C.$$
Choosing $f^{-1}(x)=\ln x,$
$$I_{\ln}(x)=x\ln x-x+C.$$
