Integral of 1/(1+x^2)

In this article we're gonna talk about how to find the indefinite integral of 1 over 1 plus x squared dx. Now in order to find the integral, we need to use trigonometric substitution. So, let's begin by replacing x with the tangent of u. So, dx is going to be the derivative of tangent which is sec squared, and it's gonna be sec squared udu. So, let's replace x with tan u, and let’s replace the dx with sec squared udu. So, we're gonna have the integral of 1 over 1 plus tangent squared which times sec squared du.

You need to be familiar with the Pythagorean identities, one of which says: 1 plus tan squared is sec squared. So we have the integral of 1 over sec squared times sec squared udu. Sec squared divided by sec squared is gonna be 1. So now we have the integral of 1 du. The antiderivative of 1 du is just gonna be u plus some constant of integration c. If x is equal to tangent u, what does u equal to? So u is going to be the inverse tangent or the arctangent of x. So the final answer is going to be the arctangent of x plus c. So, this is the integral of 1 over 1 plus x squared.

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