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Learn for free on Brilliant for a full 30 days: https://brilliant.org/polymathematic/ . You’ll also get 20% off an annual Premium subscription. There's a saying in mathematics: finding derivatives is a science, but taking integrals is an art. And if you've ever sat down with a set of integrals that look almost identical and gotten wildly different antiderivatives, you know exactly what that means. Consider three integrals, each of the form 1/(x² - 4x + c), where c is 3, 4, or 5. On paper, these look like they should behave the same way. Change one constant at the end of a quadratic, and surely the same integration technique handles all three, right? Not even close. When c is 4, the denominator is a perfect square trinomial. That means x² - 4x + 4 factors neatly into (x - 2)², and from there it's a straightforward application of the power rule for integration. Rewrite the integrand as (x - 2)⁻², bump the exponent up by one, divide — done. You get -1/(x - 2) + C. Bump that constant to 5, and suddenly you can't factor the denominator anymore. But you can complete the square: x² - 4x + 5 becomes (x - 2)² + 1. That "plus 1" is the key, because now you're looking at something that matches the derivative of inverse tangent. The antiderivative is arctan(x - 2) + C — a completely different function, arrived at through a completely different method. Drop the constant to 3, and you get yet another situation. Now x² - 4x + 3 factors into (x - 3)(x - 1), which means partial fraction decomposition is the move. Split 1/[(x - 3)(x - 1)] into A/(x - 3) + B/(x - 1), solve the system, and you end up integrating two simple reciprocals. The result involves natural logarithms: ½ ln|x - 3| - ½ ln|x - 1| + C. What makes this genuinely interesting is that the graphical behavior explains the algebraic divergence. When c = 4, there's a single vertical asymptote at x = 2. Drop to c = 3, and two asymptotes appear with a region dipping below the x-axis. Raise to c = 5, and the asymptote vanishes entirely — you get a smooth, bounded curve. Three functions that look nearly identical in symbolic form produce curves with fundamentally different geometric character. Once you see that, it stops being surprising that the antiderivatives look so different from one another. #integration #calculus #antiderivatives Watch more Math Videos: Math Minis: • Math Mini Math Minutes: • Math Minutes Number Sense: • Number Sense (UIL / PSIA) MATHCOUNTS: • MATHCOUNTS Follow Tim Ricchuiti: TikTok: / polymathematic Mathstodon: https://mathstodon.xyz/@polymathematic Instagram: / polymathematicnet Reddit: / polymath-matic Facebook: / polymathematic This video was sponsored by Brilliant.