The integration of 1/x is equal to ln|x|+C where ln denotes the natural logarithm, that is, ln|x|=log_{e}|x| and C is the integration constant. In this post, we will learn how to integrate 1/x.

## Integration of 1/x Formula

The formula for the integration of 1/x (1 divided by x) is given below.

∫ 1/x dx = ln|x| +C where ln x =log_{e}x.

## How to Integrate 1/x

Now, we will prove that ∫ 1/x dx = ln|x| +C. Note that 1/x=x^{-1} cannot be integrated using the power rule integration ∫x^{n} dx = x^{n+1}/(n+1) +C as we have -1+1=0 in the denominator. Thus, to find the integration of 1/x, we will proceed as follows:

Let z=ln x

Differentiating both sides, dz= dx/x. Therefore, we obtain that

∫1/x dx = ∫dx/x = ∫dz

= z+C

= ln x +C

= ln|x|+C as ln is not defined for negative values.

So the integration of 1/x is ln|x|+C where C is the indefinite integral constant.

**ALSO READ**

**Video Solution on Integration of 1/x:**

## Definite Integral of 1/x

**Question:** Find the definite integral of 1/x from 1 to 2, that is find ∫_{1}^{2} 1/x dx.

**Answer:**

As the integral of 1/x is equal to lnx, we obtain that

∫_{1}^{2} 1/x dx = [ln x]_{1}^{2}

⇒ ∫_{1}^{2} 1/x dx = ln 2 – ln 1

⇒ ∫_{1}^{2} 1/x dx = ln 2 – 0 as the value of ln1 is 0.

⇒ ∫_{1}^{2} 1/x dx = ln 2.

So the value of the definite integral of 1/x from 1 to 2 is equal to ln2.

## FAQs on Integration of 1/x

**Q1: What is the integration of 1/x?**

**Answer:** The integration of 1/x is ln|x|+C.