Integrating Exponentials and Logarithms

$e^{x}$ is its own integral

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

The area under $y = e^{x}$ from x = 0 to x = 1 is $\left[e^{x}\right]_{0}^{1} = e^{1} - e^{0} = e - 1 \approx 1.718$.

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

The curve $y = e^{x}$, 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 $e - 1 \approx 1.718$.

The gap in the power rule

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

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

ln x fills it

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

The area under $y = \frac{1}{x}$ from 1 to 4 is $\ln 4 - \ln 1 = \ln 4 \approx 1.386$, 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 $y = \frac{1}{x}$, measured from 1, reaches 1.

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

Areas that add when lengths multiply

The area under $\frac{1}{x}$ 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 $\frac{1}{{2x}}$ is half of $\frac{1}{x}$. 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 $y = \frac{1}{x}$, 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 $\frac{1}{x}$ does. There ln(−x) works instead: by the chain rule it differentiates to $\frac{{-1}}{{-x}} = \frac{1}{x}$. The two cases together are written with an absolute value: $\int \frac{1}{x} \,dx = \ln|x| + C$, for every x except 0.

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

A definite integral of $\frac{1}{x}$ 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 $e^{ax}$ did: $\int \frac{1}{{2x + 1}} \,dx = \frac{1}{2} \ln|2x + 1| + C$. The integral from 0 to 1 is $\frac{1}{2} \ln 3 - \frac{1}{2} \ln 1 = \frac{1}{2} \ln 3 \approx 0.549$. A constant on top stays outside: $\int \frac{5}{x} \,dx = 5 \ln|x| + C$.

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

ln x itself is integrated by parts, with ln x as the part to differentiate and 1 as the part to integrate: $\int \ln x \,dx = x \ln x - x + C$. 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 $\frac{1}{x}$. It gives $x^{0} / 0$, which has no meaning.

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

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

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

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