Solve ∫arctan(1+e^x)dx: Feynman's Trick Tutorial

by Aria Freeman 49 views

Hey math enthusiasts! Ever stumbled upon an integral that looks like it belongs in a superhero movie – complex, intimidating, and seemingly impossible to solve with conventional methods? Well, you're not alone! Today, we're going to dive deep into a fascinating problem from the MAO Nationals 2025 Mu Integration competition: evaluating the definite integral ∫[0 to ln2] arctan(1+e^x) dx. And guess what? We're going to use a super cool technique called Feynman's trick, also known as differentiation under the integral sign, to crack this mathematical puzzle.

What's the Fuss About This Integral?

Before we jump into the solution, let's take a moment to appreciate the challenge this integral presents. We're dealing with the arctangent function, which is already a bit tricky, and it's nested with an exponential function. Directly applying standard integration techniques like u-substitution or integration by parts might lead to a dead end. This is where Feynman's trick comes to the rescue, offering an elegant and powerful way to tackle such integrals.

Enter Feynman's Trick: Our Superpower

Feynman's trick, named after the brilliant physicist Richard Feynman, is a technique that allows us to evaluate integrals by introducing a parameter and differentiating under the integral sign. Sounds fancy, right? But don't worry, we'll break it down step by step. The core idea is to transform our original integral into a function of a parameter, differentiate that function with respect to the parameter, solve the resulting (hopefully simpler) integral, and then integrate back to find the value of our original integral. It's like a mathematical magic trick!

Defining the Parameterized Integral

The first step in applying Feynman's trick is to introduce a parameter into our integral. This might seem a bit arbitrary at first, but the key is to choose a parameter that simplifies the integral when we differentiate. For our integral, a clever choice is to introduce a parameter 't' inside the arctangent function:

I(t) = ∫[0 to ln2] arctan(t * e^x) dx

Notice that when t = 1, we get back our original integral. So, our goal is to find I(1). Also, observe that when t = 0, the integral becomes I(0) = ∫[0 to ln2] arctan(0) dx = 0, which is much simpler. This is a crucial observation that will help us later.

Differentiating Under the Integral Sign

Now comes the magic! We differentiate both sides of the equation with respect to 't'. This is where the