integration of log x

The integral of the log function, $log x$, can be found using integration by parts. The integral of $\ln x$ with respect to (x) is:

\[
\int \ln x \, dx = x \ln x -x + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof

To find the integral of the natural logarithm of x, $\int \ln(x) \, dx$, we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx $
In our case, we can let $ f(x) = \ln(x) $ and $ g(x) = 1 $. Then

• $ \frac {df(x)}{dx} = \frac{1}{x} \, dx $ (since the derivative of $\ln(x)$ is $\frac{1}{x}$)
• $\int g(x) dx = \int dx = x $

Now, substitute these into the integration by parts formula:

$$
\int \ln(x) \, dx = x \ln(x) – \int x \cdot \frac{1}{x} \, dx
$$

Simplifying the integral on the right:

$$
\int \ln(x) \, dx = x \ln(x) – \int 1 \, dx
$$

$$
= x \ln(x) – x + C
$$

where $ C $ is the constant of integration. So, the integral of $\ln(x)$ with respect to $x$ is $ x \ln(x) – x + C $.

Definite Integral of log x

To evaluate a definite integral of the natural logarithm of $ x $, $\int_a^b \ln(x) \, dx$, where $ a $ and $ b $ are the limits of integration, we follow a similar process as with the indefinite integral and the we’ll apply the limits at the end.

$$
\int_a^b \ln(x) \, dx = [x \ln(x) – x]_a^b
$$

$$
= [b \ln(b) – b] – [a \ln(a) – a]
$$

$$
= b \ln(b) – b – (a \ln(a) – a)
$$

$$
= b \ln(b) – b – a \ln(a) + a
$$

This is the value of the definite integral of $ \ln(x) $ from $ a $ to $ b $. Note that this formula assumes $ a $ and $ b $ are in the domain of $ \ln(x) $, which means $ a > 0 $ and $ b > 0 $.

Integration of log 10 x

So far we talked about integration of natural log x i.e $log_e x$, now we can check what will be the integration incase of common log x. It is given as

\[
\int log_{10} x \, dx = x log_{10} x – x log_{10} e + C
\]

Integration of log a x

It is given as

\[
\int log_{a} x \, dx = x log_{a} x – x log_{a} e + C
\]

Solved Examples

Question 1

Evaluate the definite integral $\int_1^e \ln(x) \, dx$.

Solution

we know $\int \ln(x) \, dx = x \ln(x) – x + C$. We apply the limits from 1 to $e$:

$$
\int_1^e \ln(x) \, dx = [x \ln(x) – x]_1^e = [e \ln(e) – e] – [1 \ln(1) – 1]
$$

Since $\ln(e) = 1$ and $\ln(1) = 0$, this simplifies to:

$$
= [e – e] – [0 – 1] = 1
$$

So, $\int_1^e \ln(x) \, dx = 1$.

Question 2

Integrate $x \ln(x)$

Solution

We will apply the integration by parts here

$$
\begin{align} \int x \ln(x) \, dx &= \frac{x^2}{2} \ln(x) – \int \left( \frac{x^2}{2} \right) \left( \frac{1}{x} \right) dx \\ &= \frac{x^2}{2} \ln(x) – \frac{1}{2} \int x \, dx \\ &= \frac{x^2}{2} \ln(x) – \frac{1}{2} \cdot \frac{x^2}{2} \\ &= \frac{x^2}{2} \ln(x) – \frac{x^2}{4} \end{align}
$$

So, the integral of $x \ln(x)$ with respect to $x$ is $\frac{x^2}{2} \ln(x) – \frac{x^2}{4} + C$, where $C$ is the constant of integration.