Is there a different approach to evaluate $\int \ln(x)\,\mathrm{d}x?$

Question

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 $\ln(x)?$

Answer 1

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 $1$.

Answer 2

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*}

Answer 3

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:

$$\begin{align}\int\left(1+\ln x\right)\text dx&=x\ln x+C\\x+\int\ln x\;\text dx&=x\ln x + C\\\int\ln x\;\text dx&=x\left(\ln x - 1\right)+C\end{align}$$

Answer 4

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

• $a’(x)=1\implies a(x)=x$

• $b’(x)=-1\implies b(x)=-x$

Answer 5

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, $$\int \ln x \; dx= x \ln x -x+C.$$

Answer 6

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:

$\displaystyle \int ye^y dy =$

$\displaystyle \int \dfrac{d}{dy}(ye^y) - \displaystyle \int e^y dy =$

$ye^y -e^y = x \ln x - x +C$.

Answer 7

Set $x = e^y$ and $y= \ln x$.

$\int \ln x \, dx = \int y \, dx = \int y \frac{dx}{dy}dy = \int y e^y dy = y e^y - e^y + C= x \ln x -x + C$

Answer 8

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.

Answer 9

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 $$\int \ln x \mathrm{d} x= x \ln x-x+C. $$

Answer 10

$$\int \ln x dx\tag1 $$

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 $y= \ln x$

$$\int ye^y dy=x\ln x-x+1\tag8$$

$$\int \ln x dx=x\ln x-x+C\tag9$$

Answer 11

$$\color{red}{\int \ln (x) \, dx}=\\\underset{n\to 0}{\text{lim}}\int \left(-\frac{1}{n}+\frac{x^n}{n}\right) \, dx=\\\underset{n\to 0}{\text{lim}}\left(-\frac{x}{n}+\frac{x^{1+n}}{n (1+n)}\right)+C=\\\underset{n\to 0}{\text{lim}}\frac{-x-n x+x^{1+n}}{n (1+n)}+C=\\\frac{\underset{n\to 0}{\text{lim}}\frac{-x+x^{1+n}-x n}{n}}{\underset{n\to 0}{\text{lim}}(1+n)}+C=\\\underset{n\to 0}{\text{lim}}\frac{-x+x^{1+n}-x n}{n}+C=\\\underset{n\to 0}{\text{lim}}\left(-x+x^{1+n} \ln (x)\right)+C=\color{red}{\\\ x \ln (x)-x+C}$$