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.

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