What is the integral of the function 1/(1+x^2)?

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The integral of the function 1/(1+x2) is a classic result in calculus. To find this integral, we can denote it as follows:

∫ (1/(1+x2)) dx

This integral is associated with the arctangent function. Thus, the solution is:

∫ (1/(1+x2)) dx = arctan(x) + C

Where C is the constant of integration. The reason this integral equals arctan(x) is rooted in the derivative of the arctangent function. To grasp this, let’s review:

If we differentiate arctan(x), we obtain:

d(arctan(x))/dx = 1/(1+x2)

This confirms that the integral is correctly identified. To summarize:

  • The integral of 1/(1+x2) is arctan(x) + C.
  • This integral is significant in many areas of mathematics, particularly in calculus and trigonometry.

In conclusion, if you encounter the integral of 1/(1+x2), you can confidently respond that it yields the result of arctan(x) + C.

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