Integrating Exponentials and Logarithms

$e^x$eˣ is its own integral

$e^x$eˣ differentiates to itself, so it integrates to itself: $\int e^{x}dx=e^x+C$∫ eˣ dx = eˣ + C. Check: $e^x+C$eˣ + C differentiates to $e^x$eˣ.

The area under $y=e^x$y = eˣ from x = 0 to x = 1 is ${\left[e^x\right]}_0^1=e^1-e^0=e-1\approx 1.718$[eˣ]₀¹ = e¹ − e⁰ = e − 1 ≈ 1.718.

With a multiple of x in the power, divide by it. $e^{3x}$e^(3x) differentiates to $3e^{3x}$3e^(3x) by the chain rule, so $\int e^{3x}dx=\frac{e^{3x}}{3}+C$∫ e^(3x) dx = e^(3x)/3 + C, and in general $\int e^{ax}dx=\frac{e^{ax}}{a}+C$∫ e^(ax) dx = e^(ax)/a + C for any constant a other than 0. For a = −1 that gives $\int e^{-x}dx=-e^{-x}+C$∫ e^(−x) dx = −e^(−x) + C. The integral of $e^{2x}$e^(2x) from 0 to 1 is $\frac{e^2-1}{2}\approx 3.195$(e² − 1)/2 ≈ 3.195.

The curve $y=e^x$y = eˣ, shaded from x = 0 to x = 1, with the tangent at (1, e). The tangent is y = ex, whose gradient e equals the height of the curve there. The shaded area is $e-1\approx 1.718$e − 1 ≈ 1.718.

The gap in the power rule

The power rule integrates $x^n$xⁿ to $\frac{x^{n+1}}{n+1}+C$xⁿ⁺¹/(n + 1) + C. For $\frac{1}{x}$1/x, which is $x^{-1}$x⁻¹, it would give $x^0/0$x⁰ / 0: adding one to −1 makes 0, and there is no dividing by 0.

No power of x can fill the gap. A power $x^k$xᵏ differentiates to $k{x}^{k-1}$k xᵏ⁻¹, and to land on $x^{-1}$x⁻¹ it would need k = 0, which makes the multiplier in front 0 as well. Yet the region under $y=\frac{1}{x}$y = 1/x from 1 to 4 plainly has an area, so $\frac{1}{x}$1/x does have an antiderivative. It is not a power.

ln x fills it

The natural logarithm differentiates to $\frac{1}{x}$1/x, so for x > 0, $\int \frac{1}{x}dx=\ln x+C$∫ 1/x dx = ln x + C.

The area under $y=\frac{1}{x}$y = 1/x from 1 to 4 is $\ln 4-\ln 1=\ln 4\approx 1.386$ln 4 − ln 1 = ln 4 ≈ 1.386, since ln 1 = 0. From 1 to e it is ln e − ln 1 = 1 exactly, which is one way to define e: the number at which the area under $y=\frac{1}{x}$y = 1/x, measured from 1, reaches 1.

The curve $y=\frac{1}{x}$y = 1/x, shaded from x = 1, where its height is 1, to x = 4, where it is 0.25. The shaded area is $\ln 4\approx 1.386$ln 4 ≈ 1.386.

Areas that add when lengths multiply

The area under $\frac{1}{x}$1/x behaves like a logarithm. The strip from 2 to 4 is the strip from 1 to 2 made twice as wide and half as tall: each point x of the first strip moves to 2x, where the height $\frac{1}{2x}$1/(2x) is half of $\frac{1}{x}$1/x. Stretching by 2 one way and squashing by 2 the other leaves the area unchanged.

So the area from 1 to 4 is the area from 1 to 2 twice, and ln 4 = 2 ln 2. The same argument works for any a and b: the area from 1 to ab is the area from 1 to a plus the area from 1 to b, so ln(ab) = ln a + ln b.

Under $y=\frac{1}{x}$y = 1/x, the gold region runs from 1 to a = 1.5 and has area ln 1.5 = 0.405. The dashed outline runs from 1 to b = 2 and has area ln 2 = 0.693. The green region runs from a to ab = 3: it is the dashed one stretched 1.5 times wider and 1.5 times shorter, so it has the same area. Gold and green together run from 1 to 3, and ln 3 = 1.099 = 0.405 + 0.693. Drag a and b so that ab = 6.

Negative x and the absolute value

For x < 0, ln x has no value, but $\frac{1}{x}$1/x does. There ln(−x) works instead: by the chain rule it differentiates to $\frac{-1}{-x}=\frac{1}{x}$(−1)/(−x) = 1/x. The two cases together are written with an absolute value: $\int \frac{1}{x}dx=\ln |x|+C$∫ 1/x dx = ln|x| + C, for every x except 0.

So the integral of $\frac{1}{x}$1/x from −4 to −1 is $\ln 1-\ln 4=-\ln 4\approx -1.386$ln 1 − ln 4 = −ln 4 ≈ −1.386, negative because the curve is below the axis there.

A definite integral of $\frac{1}{x}$1/x must not cross 0. Between −1 and 1 the curve has no value at 0, and the region on each side of 0 is infinitely large. Putting the limits into ln|x| would give ln 1 − ln 1 = 0, a number with no meaning here.

Related integrals

A linear bracket underneath divides by its multiplier, as $e^{ax}$e^(ax) did: $\int \frac{1}{2x+1}dx=\frac{1}{2}\ln |2x+1|+C$∫ 1/(2x + 1) dx = ½ ln|2x + 1| + C. The integral from 0 to 1 is $\frac{1}{2}\ln 3-\frac{1}{2}\ln 1=\frac{1}{2}\ln 3\approx 0.549$½ ln 3 − ½ ln 1 = ½ ln 3 ≈ 0.549. A constant on top stays outside: $\int \frac{5}{x}dx=5\ln |x|+C$∫ 5/x dx = 5 ln|x| + C.

Another base turns into e. $2^x$2ˣ differentiates to $2^x\ln 2$2ˣ ln 2, so $\int 2^{x}dx=\frac{2^x}{\ln 2}+C$∫ 2ˣ dx = 2ˣ/(ln 2) + C. The integral from 0 to 3 is $\frac{8-1}{\ln 2}\approx 10.10$(8 − 1)/(ln 2) ≈ 10.10.

ln x itself is integrated by parts, with ln x as the part to differentiate and 1 as the part to integrate: $\int \ln xdx=x\ln x-x+C$∫ ln x dx = x ln x − x + C. Check with the product rule: x ln x differentiates to ln x + 1, and taking away the derivative of x leaves ln x.

The usual mistakes

Applying the power rule to $\frac{1}{x}$1/x. It gives $x^0/0$x⁰ / 0, which has no meaning.

Applying the power rule to $e^x$eˣ. The x is in the exponent, not the base, so $\frac{e^{x+1}}{x+1}$eˣ⁺¹/(x + 1) does not differentiate back to $e^x$eˣ.

Forgetting to divide by a. $\int e^{3x}dx$∫ e^(3x) dx is $\frac{e^{3x}}{3}+C$e^(3x)/3 + C; $e^{3x}+C$e^(3x) + C differentiates to $3e^{3x}$3e^(3x).

Dropping the absolute value. ln x has no value for negative x, while $\frac{1}{x}$1/x does, so the integral is ln|x| + C.

Integrating $\frac{1}{x}$1/x across 0. The region is unbounded on both sides of 0, so the integral from −1 to 1 has no value.