What is the Integral of 1/x^2? [Solved] | Integration of 1/x^2

The integral of 1/x^2 is equal to -1/x +C where C is a constant. The integration formula of 1/x² 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/x²

Answer: The integration of 1/x2 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: ∫xⁿ dx = xⁿ⁺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/x² 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.

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Definite Integral of 1/x²

Question: Find the definite integral of 1/x² from 1 to 2, that is, find ∫₁² 1/x² dx.

Answer

As the integral of 1/x² 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 x²?

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