Logarithmic and Exponential Integrals: Reverse the Change

Why 1/x leads to ln absolute x

Integration asks for a function whose derivative matches the integrand. Since $d(\ln x)/dx=1/x$ for x>0, we already know an antiderivative of 1/x on the positive side. On the negative side, $-x>0$, and the chain rule gives $$\frac{d}{dx}\ln(-x)=\frac{-1}{-x}=\frac1x.$$

We can write both cases with an absolute value:

This formula applies on an interval that does not contain zero. For positive x, ln|x|=ln x; for negative x, ln|x|=ln(−x). The absolute value keeps the logarithm argument positive. It does not make the original integrand defined at zero.

The constant C matters because every vertical translate has the same derivative. On separate negative and positive intervals, the constants may be chosen independently; there is no requirement to join the antiderivatives through the missing point.

Do not try the usual power rule by substituting exponent−1 into $x^{n+1}/(n+1)$: its denominator would be zero. The reciprocal case is the special one that produces a logarithm.

A quick derivative check is more reliable than memorising a shape. Differentiate your proposed answer on the interval in the question. If its derivative equals 1/x and its logarithm exists there, you have an antiderivative. If the problem crosses zero, pause and inspect the interval before using any endpoint subtraction.

Spot a derivative divided by its function

A logarithm often hides in a fraction. If the numerator is the derivative of the denominator, reverse the chain rule:

Use this on an interval where f is differentiable and nonzero. A continuous nonzero f has a constant sign on that interval, so the absolute-value logarithm is well behaved there.

For a linear denominator, the inner derivative is its coefficient. Thus, for a≠0, $$\int\frac1{ax+b}\,dx=\frac1a\ln|ax+b|+C.$$ The factor 1/a compensates for the chain-rule factor a that appears when you differentiate the answer.

Sometimes the match is exact. In $\int2x/(x^2+4)\,dx$, the numerator 2 x is the derivative of x²+4, so the result is $\ln(x^2+4)+C$. Here x²+4 is always positive, so absolute-value signs are unnecessary.

Sometimes a constant adjustment is needed. In $\int x/(x^2+4)\,dx$, the numerator is half the inner derivative, giving $\tfrac12\ln(x^2+4)+C$.

Look for the pattern rather than declaring that every denominator gives a logarithm. For instance, $1/(x^2+4)$ lacks the factor x needed for this particular substitution. Recognising a useful method also means recognising when its numerator condition is missing. Finish by differentiating your result and checking the interval.

Undo an exponential derivative

The exponential e^x is its own derivative, so it is also its own antiderivative:

If its exponent is a linear expression, compensate for the inner derivative. For a≠0, $$\boxed{\int e^{ax+b}\,dx=\frac1a e^{ax+b}+C.}$$ Differentiating the right side brings down a and cancels the factor 1/a.

For a fixed exponential base b>0 with b≠1, the derivative of b^x contains ln b. Reverse that multiplier:

More generally, $\int b^{ax+c}\,dx=b^{ax+c}/(a\ln b)+C$ when a≠0 and b≠1. If either factor in that denominator is zero, return to the actual function instead of dividing by zero. For example, b=1 makes the integrand 1, whose antiderivative is x+C; a=0 makes the integrand the constant b^c.

A base between 0 and 1 is allowed. Its logarithm is negative, so an antiderivative such as $(1/2)^x/\ln(1/2)$ is negative before adding C. That does not contradict the positive integrand: the derivative of the antiderivative is positive. Function values and rates of change need not have the same sign.

Keep the two adjustments separate: one for the exponential base, and one for a changing inner exponent. A derivative check exposes either missing factor immediately.

Integrate ln x by choosing two factors

The integral of ln x does not match the reciprocal pattern. Instead, view it as a product: $\ln x\cdot1$. Integration by parts reverses the derivative of a product:

Choose the factor that becomes simpler when differentiated as u. On x>0, take $$u=\ln x,\quad dv=dx,\quad du=\frac{dx}x,\quad v=x.$$ Substitute every part before simplifying:

Hence $$\boxed{\int\ln x\,dx=x\ln x-x+C,\qquad x>0.}$$

Check it: the derivative of x ln x is ln x + 1, and the derivative of−x is−1. The unwanted constants cancel, leaving ln x. The subtraction is doing real work; x ln x alone is not an antiderivative.

The same choice of u helps with a polynomial times ln x. For $\int x\ln x\,dx$, keep u=ln x but choose dv=x dx, so v=x²/2. The remaining integral is simpler because the 1/x from du removes one power of x.

Before calculating, write a small two-column plan: what will be differentiated, and what will be integrated? If the second integral becomes harder, reconsider that choice. Here the logarithm is the factor to differentiate because its derivative is a simple reciprocal.

Add an interval before subtracting endpoints

A definite integral combines an integrand with an interval. If 0<a<b, the function 1/x is positive and continuous on[a,b], so $$\int_a^b\frac{dx}x=\ln b-\ln a=\ln\left(\frac ba\right).$$ Its value is also the area below 1/x between those positive bounds. Reversing the bounds reverses the sign of the integral.

The answer depends on a ratio. From 1 to 2 and from 2 to 4, the integral is the same: ln 2. The second interval is twice as wide, but its curve is correspondingly lower. Use the area exploration to compare these intervals without confusing width with area.

Now inspect $\int_{-1}^{1}1/x\,dx$. Zero lies inside the interval, and 1/x is undefined there. You cannot substitute endpoints into ln|x| and announce 0. The fundamental endpoint rule needs its hypotheses: this integrand is not continuous across the interval.

If interpreted as an improper integral, the negative and positive sides must each converge separately. They do not, so the integral does not converge. Equal-looking areas on a symmetric sketch do not justify cancelling two divergences.

For every definite integral, use this order: check the whole interval, choose an antiderivative valid there, evaluate the upper endpoint, then subtract the lower endpoint. A logarithm’s absolute value keeps an argument positive but cannot repair a singularity inside the integration interval.