JEE Main Integration Practice Test

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Question: 1 / 400

What is the integral ∫ (1/x^2) dx?

-1/x + C

To find the integral of $ \int \frac{1}{x^2} \, dx $, we can rewrite the integrand using the power rule. The function $ \frac{1}{x^2} $ can be expressed as $ x^{-2} $.

Using the power rule of integration, which states that $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $ for any real number $ n \neq -1 $, we can apply it here. Here, $ n = -2 $.

Applying the power rule:

\int x^{-2} \, dx = \frac{x^{-2 + 1}}{-2 + 1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C

Thus, the integral evaluates to $ -\frac{1}{x} + C $, which is why the correct answer is indeed the first option.

The constant $ C $ represents the constant of integration, which arises because the operation of integration can yield multiple functions that differ by a constant. Therefore, \( -\frac

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1/x + C

2/x + C

-2/x + C

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