The Magic of One Integral: Four Beautiful Ways to See It

What do you get when you combine algebra, calculus, infinite series, and a hint of complex numbers? You get a beautiful little integral that opens a whole world of connections.

Today, we’ll explore the value of the following definite integral:

∫01ln⁡(1+x)1+x2 dx\int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx

We’ll solve it four different ways — each drawing from a different area of mathematics — and by the end, you’ll see how deep even a single problem can go.

Method 1: Infinite Series

We start by expanding the natural logarithm using its Taylor series:

ln⁡(1+x)=∑n=1∞(−1)n+1xnn\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}

This series is valid for x∈(−1,1]x \in (-1, 1], and converges nicely on our interval [0,1][0, 1]. Now plug this into the integral:

∫01ln⁡(1+x)1+x2 dx=∑n=1∞(−1)n+1n∫01xn1+x2 dx\int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \int_0^1 \frac{x^n}{1+x^2} \, dx

Each of these inner integrals can be computed individually — for example:

  • When n=1n = 1: ∫01x1+x2dx=12ln⁡(2)\int_0^1 \frac{x}{1+x^2} dx = \frac{1}{2} \ln(2)

  • When n=2n = 2: Use algebra to write x21+x2=1−11+x2\frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}, which gives another exact value.

Adding a few terms gives a great approximation, and with work, you find:

∫01ln⁡(1+x)1+x2 dx=π8ln⁡(2)\int_0^1 \frac{\ln(1+x)}{1+x^2} \, dx = \frac{\pi}{8} \ln(2)

Method 2: Trigonometric Substitution

Let’s try a change of variables: set x=tan⁡θx = \tan \theta. Then we get:

  • dx=sec⁡2θ dθdx = \sec^2 \theta \, d\theta

  • 1+x2=sec⁡2θ1 + x^2 = \sec^2 \theta

This simplifies our integral:

∫01ln⁡(1+x)1+x2dx=∫0π4ln⁡(1+tan⁡θ) dθ\int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) \, d\theta

Now here's the clever part. Replace θ\theta with π4−θ\frac{\pi}{4} - \theta. That gives:

∫0π4ln⁡(1+tan⁡θ) dθ=∫0π4ln⁡(1+cot⁡θ) dθ\int_0^{\frac{\pi}{4}} \ln(1 + \tan \theta) \, d\theta = \int_0^{\frac{\pi}{4}} \ln(1 + \cot \theta) \, d\theta

Add these two:

2I=∫0π4ln⁡((1+tan⁡θ)(1+cot⁡θ))dθ2I = \int_0^{\frac{\pi}{4}} \ln\left((1 + \tan \theta)(1 + \cot \theta)\right) d\theta

With a trig identity, this simplifies to a function that integrates nicely — and again, we arrive at:

∫01ln⁡(1+x)1+x2dx=π8ln⁡(2)\int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \frac{\pi}{8} \ln(2)

Method 3: Feynman's Trick (Differentiation Under the Integral Sign)

Richard Feynman, the famous physicist, used this trick all the time. We define a function with a parameter:

F(α)=∫01ln⁡(1+αx)1+x2dxF(\alpha) = \int_0^1 \frac{\ln(1+\alpha x)}{1+x^2} dx

Notice that F(1)F(1) is our original integral. We differentiate F(α)F(\alpha):

F′(α)=∫01x(1+x2)(1+αx)dxF'(\alpha) = \int_0^1 \frac{x}{(1+x^2)(1+\alpha x)} dx

Then use partial fractions to break the integrand down and integrate it easily. Finally, integrate F′(α)F'(\alpha) back from α=0\alpha = 0 to α=1\alpha = 1 to recover F(1)F(1). This process gives us:

F(1)=∫01ln⁡(1+x)1+x2dx=π8ln⁡(2)F(1) = \int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \frac{\pi}{8} \ln(2)

No tricks — just smooth calculus magic.

Method 4: Complex Analysis (Advanced but Awesome)

This one's for the math enthusiasts!

We consider the complex-valued function:

f(z)=log⁡(1+z)1+z2f(z) = \frac{\log(1 + z)}{1 + z^2}

We integrate this function around a contour in the complex plane that loops around the singularity at z=iz = i. Using the residue theorem from complex analysis, we compute:

∫−∞∞log⁡(1+x)1+x2dx=πlog⁡(1+i)\int_{-\infty}^\infty \frac{\log(1 + x)}{1 + x^2} dx = \pi \log(1 + i)

Now, log⁡(1+i)=ln⁡(2)+iπ4\log(1 + i) = \ln(\sqrt{2}) + i\frac{\pi}{4}, so:

Re⁡(∫−∞∞log⁡(1+x)1+x2dx)=πln⁡(2)=π2ln⁡(2)\operatorname{Re} \left( \int_{-\infty}^\infty \frac{\log(1 + x)}{1 + x^2} dx \right) = \pi \ln(\sqrt{2}) = \frac{\pi}{2} \ln(2)

The function is real and well-behaved on [0,1][0, 1], and it turns out — using symmetry and change-of-variable arguments — that the area under ln⁡(1+x)1+x2\frac{\ln(1+x)}{1+x^2} from 00 to 11 is exactly one-fourth of the real part of the full integral over (−∞,∞)(-\infty, \infty).

So again:

∫01ln⁡(1+x)1+x2dx=14⋅π2ln⁡(2)=π8ln⁡(2)\int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \frac{1}{4} \cdot \frac{\pi}{2} \ln(2) = \frac{\pi}{8} \ln(2)

Why This Is Cool

Here’s what we used:

  • Series expansions from calculus

  • Trig substitutions from integration techniques

  • Feynman's trick, which gives a new way to think about parameters

  • A taste of complex analysis, the math behind electrical engineering, physics, and quantum mechanics!

And they all point to the same beautiful result:

∫01ln⁡(1+x)1+x2dx=π8ln⁡(2)\int_0^1 \frac{\ln(1+x)}{1+x^2} dx = \frac{\pi}{8} \ln(2)

The Big Takeaway

Math isn't just about finding answers — it's about finding connections. One integral led us through series, calculus, symmetry, and even the complex plane. And whether you're a student or a mathematician, that kind of unity is what makes math magical.

Curious to try your own? Try evaluating: ∫01ln⁡(1−x)1+x2dx\int_0^1 \frac{\ln(1 - x)}{1 + x^2} dx

(You'll need to adjust the branch of the logarithm and might find some symmetry too!)

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