# How to Find the Antiderivative of Ln(x)? Nope, it's Not Difficult

Though it may be basic, computing the antiderivative of ln(x) requires the use of some important, fundamental concepts of calculus. In this Buzzle post, we tell you what these concepts are, and how you can use them to find the antiderivative of ln(x).

ScienceStruck Staff

Last Updated: Oct 29, 2018

Did You Know?

The word 'calculus' is a Latin word, which means a small stone used for counting. Today, calculus refers to any method or system of calculation that comprises symbolic manipulation of the different expressions.

Integral of Ln(x) by Parts

**Step 1**

We will start by using the product rule of derivatives. Consider that there are two functions - f(x) and g(x), and we have to find the derivative of the multiplication of those two. This is represented as following.

**((f(x) · g(x))' = f(x) · g'(x) + f'(x) · g(x) -------------(1)**

**Step 2**

Now, we will integrate both the sides of equation (1).

**∫ ((f(x) · g(x))' = ∫ { f(x) · g'(x) + f'(x) · g(x) }**

f(x) · g(x) = ∫ f(x) · g'(x) dx + ∫ f'(x) · g(x) dx ---------------- (2)

f(x) · g(x) = ∫ f(x) · g'(x) dx + ∫ f'(x) · g(x) dx ---------------- (2)

*Note: Here, we don't need to include an integration constant on the left side of the equation, as the integrals on the right side already have integration constants.*

**Step 3**

Rearranging the terms in equation (2), we get,

**∫ f(x) · g'(x) dx = f(x) · g(x) - ∫ f'(x) · g(x) dx**

The equation obtained in Step 3 is the formula for computing integration by parts. It is frequently used for solving difficult integration problems.

**f(x) = u**

g(x)'dx = dv

f'(x)dx = du

g(x) = v

g(x)'dx = dv

f'(x)dx = du

g(x) = v

**∫ udv = uv - ∫ vdu**

To integrate by parts, keep in mind to choose 'u' and 'dv' strategically, so that the calculation becomes easy.

How to Find the Antiderivative of Ln(x)

**∫ ln(x) dx**

**∫ ln(x) · 1dx**

**u = ln(x) ----------(a)**

and,

**dv = 1 · dx -----------(b)**

Taking the derivative of u, we get,

**du = (1/x) dx ------------(c)**

**∫ du = ∫ dv**

u = x ------------(d)

u = x ------------(d)

We have the formula for integration by parts as,

**∫ udv = uv - ∫ vdu**

**∫ ln(x) dx = ln(x) · x dx - ∫ 1/x · x dx**

= x * ln(x) - ∫ dx

= x * ln(x) - x + C

= x * ln(x) - ∫ dx

= x * ln(x) - x + C

**[x * ln(x) - x + C]' = ln(x) + (x/x) - 1 + 0**

= ln(x) + 1 - 1 + 0

= ln(x)

= ln(x) + 1 - 1 + 0

= ln(x)