Is there a different approach to evaluate $\int \ln(x)\,\mathrm{d}x?$ The usual method of evaluating $\int \ln(x)\,\mathrm{d}x$ requires you to rewrite it as 
$$
\int \ln(x) \cdot \color{brown}1\,\mathrm{d}x
$$
and apply integration by parts.
Letting $u=\ln(x)$ and $\mathrm{d}v=\color{brown}1 \,\mathrm{d}x$, we get that $\mathrm{d}u=\frac{1}{x} \,\mathrm{d}x$ and $v=\int \color{brown}1 \,\mathrm{d}x = x$, so our integral becomes
\begin{align}
\int \ln(x) \cdot \color{brown}1\,\mathrm{d}x
&=uv-\int v\,\mathrm{d}u
\\[0.5em]
&=x\ln(x)-\int x\cdot\frac{1}{x}\,\mathrm{d}x
\\[0.5em]
&=x\ln(x)-\int \mathrm{d}x
\\[0.5em]
&=x\ln(x)-x+c
\end{align}
After looking through many calculus books, this is the only method I've found to integrate $\ln(x)$. Are there any other methods one could use to integrate the function $\ln(x)?$
 A: You could always eliminate the log by substituting $x=e^u$.  We have
$$\int\ln x\,dx=\int ue^u\,du\ .$$
Now you will still need integration by parts, but in this case it's a pretty obvious integration by parts which does not rely on the "trick" of inserting a factor of $1$.
A: By trial solution 
$$F(x)=a(x)\log x+b(x) \implies F’(x)=a’(x) \log x+\frac{a(x)}{x}+b’(x)$$
then


*

*$a’(x)=1\implies a(x)=x$

*$b’(x)=-1\implies b(x)=-x$
A: Rough reasoning:
\begin{align*}
\ln x&=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}(x-1)^{n}\\
\int\ln xdx&=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}\cdot\dfrac{1}{n+1}(x-1)^{n+1}\\
&=\sum_{n=1}^{\infty}(-1)^{n+1}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)(x-1)^{n+1}\\
&=(x-1)\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}(x-1)^{n}+\sum_{n=1}^{\infty}(-1)^{n+2}\dfrac{1}{n+1}(x-1)^{n+1}\\
&=(x-1)\ln x+\ln x-(x-1)\\
&=x\ln x-x+c.
\end{align*}
A: Since for $t>1$ (the other case is similar) $\int_1^t \ln x dx$ is the (positive) area of the region $D\subset \mathbb R^2$ enclosed by the curve $y=\ln x$, the straight line $x=t$ and the horizontal axis, this area can also be written as a double integral $\iint_D dA$, which by Fubini theorem equals the iterated integral
$$\int_1^t \int_0^{\ln x} dy\, dx.$$
Changing order of integration by writing $D=\{(x,y)\in\mathbb R^2 \colon 0\le y\le \ln t \wedge e^y \le x \le t\}$, that integral is equal to
$$\int_0^{\ln t} \int_{e^y}^t dx\,dy=\int_0^{\ln t} (t-e^y) \;dy=(ty-e^y)|_{y=0}^{\ln t}=t \ln t -t+1.$$
That is,
$$\int \ln x \; dx= x \ln x -x+C.$$
A: I'm not sure that this is what you're looking for, but we can "discover" $\int\ln x\;\text dx$ by differentiating the function $x\ln x$.
By using the product rule we get:
$$\frac{\text d}{\text dx}\left(x\ln x\right)=1 + \ln x$$
This means that:
$$\begin{align}\int\left(1+\ln x\right)\text dx&=x\ln x+C\\x+\int\ln x\;\text dx&=x\ln x + C\\\int\ln x\;\text dx&=x\left(\ln x - 1\right)+C\end{align}$$
A: For fun, a bit of trickery:
Let $x=e^y$.
Then:
$\displaystyle \int ln x dx = \displaystyle \int ye^y dy$;
Now look at :
$ \dfrac{d}{dy} (ye^y) = e^y +ye^y$.
Thus:
$\displaystyle \int ye^y dy =$
$\displaystyle \int \dfrac{d}{dy}(ye^y) - \displaystyle \int e^y dy =$
$ye^y -e^y =  x \ln x - x +C$.
A: Set $x = e^y$ and $y= \ln x$.
$\int \ln x \, dx = \int y \, dx = \int y \frac{dx}{dy}dy = \int y e^y dy = y e^y - e^y + C=  x \ln x -x  + C$
A: $$\int \ln x dx\tag1 $$
Let $y=\ln x$ then ${dy\over dx}={1\over x}$, $dx=xdy$
and $x=e^y$
$$\int yxdy=\int ye^ydy\tag2$$
now using the series of $e^y$: $$e^y=1+y+{y^2\over 2!}+{y^3\over 3!}+\cdots\tag3$$ 
$$\int ye^ydy=\int \left(y+y^2+{y^2\over 2!}+{y^3\over  3!}+\cdots\right)dy\tag4$$
$$\int ye^ydy={y^2\over 2}+{y^3\over 3\cdot 1}+{y^4\over 4\cdot 2!}+\cdots=\sum_{n=0}^{\infty}{y^{n+2}\over (n+2)n!}=y\sum_{n=0}^{\infty}{y^{n+1}\over (n+1)!}-\sum_{n=0}^{\infty}{y^{n+2}\over (n+2)!}\tag5$$
$$\int ye^ydy=y\sum_{n=0}^{\infty}{y^{n+1}\over (n+1)!}-\sum_{n=0}^{\infty}{y^{n+2}\over (n+2)!}=y(e^y-1)-(e^y-1-y)\tag6$$
Simplify
$$\int ye^y dy=y(e^y-1)-(e^y-1-y)=ye^y-e^y+1\tag7$$
$e^y=x$ and $y= \ln x$
$$\int ye^y dy=x\ln x-x+1\tag8$$
$$\int \ln x dx=x\ln x-x+C\tag9$$
A: $$\color{red}{\int \ln (x) \, dx}=\\\underset{n\to 0}{\text{lim}}\int \left(-\frac{1}{n}+\frac{x^n}{n}\right) \, dx=\\\underset{n\to
   0}{\text{lim}}\left(-\frac{x}{n}+\frac{x^{1+n}}{n (1+n)}\right)+C=\\\underset{n\to 0}{\text{lim}}\frac{-x-n x+x^{1+n}}{n
   (1+n)}+C=\\\frac{\underset{n\to 0}{\text{lim}}\frac{-x+x^{1+n}-x n}{n}}{\underset{n\to 0}{\text{lim}}(1+n)}+C=\\\underset{n\to
   0}{\text{lim}}\frac{-x+x^{1+n}-x n}{n}+C=\\\underset{n\to 0}{\text{lim}}\left(-x+x^{1+n} \ln (x)\right)+C=\color{red}{\\\ x \ln (x)-x+C}$$
A: Suppose that $F(x)$ is an antiderivative of $\log x$; that is to say, $F'(x) = \log x$.  Then by the Fundamental Theorem, $$\int_{x=1}^c \log x \, dx = F(c) - F(1).$$  But the LHS is simply the area under the curve of $\log x$ from $x = 1$ to $x = c$.  If $c > 1$, this is also the area below the line $y = c$ but above the curve $y = e^x$ from $x = 0$ to $x = \log c$; that is to say, $$\int_{x=1}^c \log x \, dx = \int_{x=0}^{\log c} c - e^x \, dx = \Bigl[ cx - e^x \Bigr]_{x=0}^{\log c} = (c \log c - e^{\log c}) - (0 - 1) = c \log c - c + 1.$$  Consequently we may take $F(c) = c \log c - c$.
If $0 < c < 1$, then the area corresponds instead to $$-\int_{x=\log c}^0 e^x - c \, dx,$$ which of course yields the same result.
A: The fundamental theorem of calculus implies that
$$ f(x)=\int_1 ^x \ln u \mathrm{d} u $$
gives an anti-derivative, and all others differ from this one by an additive constant. We further have
$$f(x)=\int_1^x \int_1^u \frac{1}{v} \mathrm{d}v \mathrm{d} u, $$
and using the Fubini-Tonelli theorem gives
$$f(x)=\int_1^x \int_v^x \frac{1}{v} \mathrm du \mathrm{d} v=\int_1^x  \left(\frac{x}{v}-1 \right) \mathrm{d} v=x \ln x-(x-1). $$
It follows that the indefinite integral
$$\int \ln x \mathrm{d} x= x \ln x-x+C. $$
