Three families of derivative, exponential, logarithmic, and inverse trig, show up constantly for the rest of this course. They almost always appear in reverse: recognizing an integrand as ONE of these derivatives, rather than computing the derivative itself. This section is a fast, focused pass on the recognition direction, since that's the direction integration needs.
\dfrac{1}{1+x^2} looks like nothing special algebraically. No obvious antiderivative jumps out from the power rule or substitution. But it's secretly a DERIVATIVE in disguise, the same way 2x\cos(x^2) was a chain-rule derivative in disguise two sections ago. Once \arctan x's derivative is memorized, spotting its shape inside an integral is the entire trick. No new technique is required, just recognition. The same habit applies to all three families at once: spot the derivative shape first, and the antiderivative follows for free.
Exponentials are functions where the variable sits in the exponent, like e^x. Logarithms are their inverse, undoing exactly what an exponential does. Inverse trig functions are the functions that undo a trig function, the way \arctan undoes tangent. All three families' derivatives show up constantly for the rest of this course. They almost always appear in reverse: recognizing an integrand as ONE of these derivatives, rather than computing the derivative itself. \frac{d}{dx}[e^x] = e^x makes e^x its own antiderivative, the only elementary function with that property. \frac{d}{dx}[\ln x] = \frac{1}{x} runs the other way. It's the reason \int \frac{1}{x}\,dx = \ln|x|+C closes a gap the power rule can't close. x^{-1} is the one power of x where "raise the exponent, divide by it" fails outright, since the new exponent would be 0. Inverse trig derivatives close a different gap. \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2} means an integrand shaped like \frac{1}{1+x^2} integrates to \arctan x. That shape is not a power and not a chain-rule form, and it's a fact with no other route to it.
e to the x meets its own slope
Set a — notice a*e^x and its derivative are always the SAME curve, just scaled.
Loading interactive…
Exponentials and logs really do undo each other
Drag the probe -- e^{\ln x} always lands back on x itself, for every x>0.
Loading interactive…
Worked example
Recognizing an inverse-trig antiderivative
-
1
Evaluate \displaystyle\int \frac{dx}{\sqrt{4-x^2}}.
-
2memorized shape, not derived fresh each time
This matches the shape \dfrac{1}{\sqrt{a^2-x^2}} with a=2, whose antiderivative is \arcsin\!\left(\dfrac{x}{a}\right).
memorized shape, not derived fresh each time
-
3
Substitute a=2 directly: \boxed{\int \frac{dx}{\sqrt{4-x^2}} = \arcsin\!\left(\dfrac{x}{2}\right) + C}
Loading interactive…
Where this goes next
Every remaining chapter
e^x, \ln x, and the inverse trig functions reappear constantly. Exponential growth/decay models, logarithmic integrals from partial fractions, and inverse-trig antiderivatives from trig substitution all lean on these shapes being recognized on sight.
Check your understanding (counts toward your progress)
Find \displaystyle\int \dfrac{dx}{\sqrt{4-x^2}}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int\frac{dx}{\sqrt{a^2-x^2}} = \arcsin\left(\frac{x}{a}\right) with a=2 — \arcsin\left(\frac{x}{2}\right), plus C.
Correct answer
Also accepted:
Why
This matches \int \dfrac{dx}{\sqrt{a^2-x^2}} with a=2.
Find \displaystyle\int \dfrac{dx}{9+x^2}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int\frac{dx}{a^2+x^2} = \frac1a\arctan\left(\frac{x}{a}\right) with a=3 — \tfrac13\arctan\left(\tfrac{x}{3}\right), plus C.
Correct answer
Also accepted:
Why
This matches \int \dfrac{dx}{a^2+x^2} with a=3.
Find \displaystyle\int_0^{1911}\dfrac{1}{1911^2+x^2}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\displaystyle\int_0^{1911}\dfrac{1}{1911^2+x^2}\,dx=\dfrac{1}{1911}\left[\arctan(1)-\arctan(0)\right]=\dfrac{1}{1911}\cdot\dfrac{\pi}{4}\approx0.0004.
Correct answer
Also accepted:
Why
\displaystyle\int\dfrac{1}{a^2+x^2}\,dx=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C.
Find \displaystyle\int_0^{878}\dfrac{1}{878^2+x^2}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\displaystyle\int_0^{878}\dfrac{1}{878^2+x^2}\,dx=\dfrac{1}{878}\left[\arctan(1)-\arctan(0)\right]=\dfrac{1}{878}\cdot\dfrac{\pi}{4}\approx0.0009.
Correct answer
Also accepted:
Why
\displaystyle\int\dfrac{1}{a^2+x^2}\,dx=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)+C.