The integral of 1/x^2 is equal to -1/x +C where C is a constant. The integration formula of 1/x^{2} is as follows:

$\int \dfrac{1}{x^2} dx = -\dfrac{1}{x}$ +C.

In this post, we will learn how to integrate 1/x^2.

## Integration of 1/x^{2}

Answer: The integration of 1/x^{2} is equal to -1/x +C. |

**Explanation:**

**Step 1:**

Using the rule of indices $\dfrac{1}{c^n}$ = c^{-n}, we have that

$\dfrac{1}{x^2}$ = x^{-2}

**Step 2:**

Now, integrating both sides we get that

$\int \dfrac{1}{x^2} dx$ = ∫x^{-2} dx

**Step 3:**

Apply the power rule of integration: ∫x^{n} dx = x^{n+1}/(n+1). Therefore,

$\int \dfrac{1}{x^2} dx =\dfrac{x^{-2+1}}{-2+1}+C$

⇒ $\int \dfrac{1}{x^2} dx =\dfrac{x^{-1}}{-1}+C$

⇒ $\int \dfrac{1}{x^2} dx = -\dfrac{1}{x}+C$

So the integration of 1/x^{2} is equal to -1/x +C where C is an arbitrary integral constant, and this is proved by the rule of indices and the power rule of integration.

**More Integrals:** Integration of x^{n}

What is the Integration of 1/x^{n}

## Definite Integral of 1/x^{2}

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

**Answer**

As the integral of 1/x^{2} is -1/x, we have that

$\int \dfrac{1}{x^2} dx$ $=\Big[-\dfrac{1}{x} \Big]_1^2$

= $-(\dfrac{1}{2} – \dfrac{1}{1} )$

= $-\dfrac{1}{2} +1$

= 1/2

So the definite integration of 1/x^2 from 1 to 2 is equal to 1/2.

**Read Also:** What is the integration of x^{2}?

## FAQs

### Q1: What is the integral of x^2?

Answer: The integral of x square is equal to x^{3}/3+C where C is an integral constant.