The integral of 1/x^n (1 by x power n) is equal to x^{-n+1}/(1-n) +C. Here we will learn how to integrate 1/x^{n} using the power rule of integration.

The integration formula of 1/x^{n} is given by

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

where C denotes an integral constant.

## Integration of 1/x^{n}

The integral of 1/x^{n } is denoted by ∫dx/x^{n}. We have:

$\int \dfrac{dx}{x^n}$

= $\int x^{-n} \, dx$ by the rule of indices.

= $\dfrac{x^{-n+1}}{1-n}+C$ by the power rule of integrations.

So the integration of 1 by x^n is equal to x^{1-n}/(1-n) +C (C is a constant), and this is obtained by the power rule of integrals.

You Can Read These Integrals:

Integral of ln(x) | Integral of sin^{2}x

Integral of cos^{2}x | Integral of sin^{3}x

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

**Question: **Find the integration of 1/x^{n} from 0 to 1, that is,

Find $\int_0^1 \dfrac{dx}{x^n}$.

**Answer:**

AS the integration of 1/x^{n} is equal to x^{-n+1}/(1-n) +C by above, we obtain that

$\int_0^1 \dfrac{dx}{x^n}$

= $\Big[\dfrac{x^{1-n}}{1-n} \Big]_0^1$

= $\dfrac{1^{1-n}}{1-n} – \dfrac{0^{1-n}}{1-n}$

= $\dfrac{1}{1-n} – 0$

= $\dfrac{1}{1-n}$.

So the integration of 1/x^{n} from 0 to 1 is equal to 1/(n-1).

**More Reading:** How to integrate 2^{x}

Integration of square root of a^{2}+x^{2}

Integration of square root of a^{2}-x^{2}

## FAQs

### Q1: What is the integration of 1/x^{n}?

Answer: The integration of 1/x^{n }is equal to x^{1-n}/(1-n) +C where C is an arbitrary integral constant.