Integration of Logarithmic Functions With Examples

What Is Integration Of Logarithmic Functions?

Integration of logarithmic functions is the process of finding antiderivatives of expressions that contain a logarithm, such as $\ln x$, $\log_a x$, $x^2\ln x$ or $(\ln x)^2$. The anchor result, valid for $x > 0$, is

$$\int \ln x,dx = x\ln x - x + C.$$

Here $\ln x$ is the natural logarithm, the logarithm with base $e \approx 2.71828$. The variable $x$ must be positive because $\ln x$ is only defined there, and $C$ is the constant of integration that every indefinite integral carries.

The topic has two directions, and textbooks often mix them under one heading. One direction integrates a function whose answer turns out to be a logarithm. The other integrates a function that already contains a logarithm.

The first direction is covered in depth on the natural-log page, so this guide spends its time on the second: the indefinite integrals of $\ln x$, $\log_a x$, $\ln(ax+b)$, $x^n\ln x$ and $(\ln x)^n$, plus their definite versions.

How Do You Integrate The Natural Logarithm By Parts?

You integrate $\ln x$ by writing it as $\ln x \cdot 1$ and applying integration by parts with $u = \ln x$ and $dv = dx$. The formula is $\int u,dv = uv - \int v,du$, and the point of the choice is that you get to differentiate the logarithm (easy) instead of integrating it (the thing you cannot yet do).

A student on Physics Forums asked exactly the right question about this setup: "where does the x come from?" It comes from $dv = dx$. If $dv = dx$, then $v = \int 1,dx = x$, and that $x$ is the one that appears in front of $\ln x$ in the answer.

$$\int \ln x,dx = \int (\ln x)(1),dx$$

$$u = \ln x, \qquad du = \frac{1}{x},dx$$

$$dv = dx, \qquad v = x$$

$$\int \ln x,dx = x\ln x - \int x \cdot \frac{1}{x},dx$$

$$= x\ln x - \int 1,dx$$

$$= x\ln x - x + C$$

Check by differentiating, using the product rule on $x\ln x$:

$$\frac{d}{dx}\big(x\ln x - x\big) = \ln x + x \cdot \frac{1}{x} - 1 = \ln x.$$

The $dv = dx$ choice felt like cheating the first time I met it; it only started to feel legitimate after the derivative check came back clean. The same lone-function trick integrates $\arctan x$ and $\arcsin x$, which is why the LIATE order (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) puts logs first when you pick $u$. The derivatives of logarithmic functions are what make $du$ so simple here.

Is There A Way To See The Integral Of The Natural Log Without Integration By Parts?

Yes: a rectangle and the exponential curve give the same formula with no integration by parts at all. Take any $b > 1$ and draw the rectangle with corners $(0, 0)$ and $(b, \ln b)$. Its area is $b\ln b$.

The curve $y = \ln x$ cuts that rectangle into two pieces:

• The piece under the curve, between $x = 1$ and $x = b$, which is the area $\int_{1}^{b} \ln x,dx$ you want.

• The piece to the left of the curve, between the $y$-axis and the curve for $y$ from $0$ to $\ln b$. Reading the curve sideways as $x = e^{y}$, this area is $\int_{0}^{\ln b} e^{y},dy = b - 1$.

Subtracting the left piece from the rectangle:

$$\int_{1}^{b} \ln x,dx = b\ln b - (b - 1) = b\ln b - b + 1.$$

That is exactly $\big[x\ln x - x\big]_{1}^{b}$, so the geometric picture and the algebra agree. Setting $b = e$ gives $e - e + 1 = 1$, the shaded area in the opening diagram. The picture works because $\ln x$ and $e^{x}$ are inverse functions, so the two pieces always fill the rectangle.

How Do You Integrate A Logarithm With Base A Or Base 10?

To integrate $\log_a x$, convert it to the natural log with the change-of-base rule, pull out the constant $\frac{1}{\ln a}$, and use the $\ln x$ result. The rule is one of the properties of logarithms:

$$\log_a x = \frac{\ln x}{\ln a}, \qquad a > 0,\ a \neq 1.$$

$$\int \log_a x,dx = \frac{1}{\ln a}\int \ln x,dx$$

$$= \frac{x\ln x - x}{\ln a} + C$$

$$= x\log_a x - \frac{x}{\ln a} + C$$

A Quora thread asked for tips on integrating "logarithmic functions like ln(x) and log(x)", and the base is the tip that matters most. On most calculators and in many school textbooks, "log" means base 10, while in many university texts and in software "log" means base $e$. For base 10 the factor is $\frac{1}{\ln 10} \approx 0.4343$, so

$$\int \log_{10} x,dx = \frac{x\ln x - x}{\ln 10} + C.$$

If a problem writes $\log x$ with no base, check the source's convention before you integrate. The two answers differ by that factor of about $0.4343$.

How Do You Integrate A Logarithm Of A Linear Expression?

For $\ln(ax + b)$, substitute $u = ax + b$ so that $du = a,dx$, then reuse the $\ln u$ result. The domain condition is $ax + b > 0$ and $a \neq 0$.

$$\int \ln(ax+b),dx = \frac{1}{a}\int \ln u,du$$

$$= \frac{1}{a}\big(u\ln u - u\big) + C$$

$$= \frac{ax+b}{a}\ln(ax+b) - x + C$$

The last line folds the constant $-\frac{b}{a}$ into $C$, which is allowed because any constant can be absorbed. When $b = 0$ there is a shortcut: $\ln(ax) = \ln a + \ln x$, so $\int \ln(ax),dx = x\ln a + x\ln x - x + C = x\ln(ax) - x + C$.

You might hope to split $\ln(2x + 3)$ the same way, as $\ln 2x + \ln 3$. Hold on, though: the product rule for logs needs a product inside the bracket, and $2x + 3$ is a sum. So $\ln(2x+3)$ does not split, and the substitution route is the only clean one.

How Do You Integrate A Power Times The Natural Log?

For $x^n\ln x$ with $n \neq -1$, use integration by parts with $u = \ln x$ and $dv = x^n,dx$. A Brainly question put it plainly, "What is the integral of ln(x)x?", and the general answer covers that case and every other power at once.

$$u = \ln x, \qquad du = \frac{1}{x},dx$$

$$dv = x^n,dx, \qquad v = \frac{x^{n+1}}{n+1}$$

$$\int x^n\ln x,dx = \frac{x^{n+1}}{n+1}\ln x - \int \frac{x^{n+1}}{n+1}\cdot\frac{1}{x},dx$$

$$= \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C$$

The excluded case $n = -1$ is $\frac{\ln x}{x}$, where $v$ would need $\frac{x^{0}}{0}$. That integrand belongs to substitution, covered two sections down.

How Do You Integrate Powers Of The Natural Log?

Powers such as $(\ln x)^2$ fall to integration by parts with $u = (\ln x)^n$ and $dv = dx$, applied repeatedly. A Vedantu question asked to "evaluate the integral of (log x)^2 dx", which is the first case of the general pattern. The parts step gives a reduction formula, an identity that lowers the power by one each time:

$$\int (\ln x)^n,dx = x(\ln x)^n - n\int (\ln x)^{n-1},dx.$$

For $n = 2$, apply it once and then use the $\ln x$ result:

$$\int (\ln x)^2,dx = x(\ln x)^2 - 2\int \ln x,dx$$

$$= x(\ln x)^2 - 2x\ln x + 2x + C$$

For $n = 3$, apply it twice:

$$\int (\ln x)^3,dx = x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln x - 6x + C$$

The coefficients $1, 3, 6, 6$ follow a falling pattern ($3$, then $3 \times 2$, then $3 \times 2 \times 1$), which is a quick way to check a long answer. The reduction-formula idea reappears for powers of sine and cosine later in an integration course.

When Does Substitution Beat Integration By Parts For Log Integrals?

Substitution wins whenever the integrand contains $\ln x$ together with a factor of $\frac{1}{x}$, because $\frac{1}{x},dx$ is exactly $d(\ln x)$. A Brainly question, "integration ln x/x dx", is the standard example. With integration by substitution and $u = \ln x$, $du = \frac{1}{x},dx$:

$$\int \frac{\ln x}{x},dx = \int u,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C.$$

The quick way to choose a method is to look for that $\frac{1}{x}$ factor first:

The last row is a warning, not a method. The integral of $\frac{1}{\ln x}$ has no formula built from elementary functions; it defines the logarithmic integral $\operatorname{li}(x)$, a special function listed in the NIST Digital Library of Mathematical Functions. Its value at $x = 10$ is about $6.1656$, found numerically.

How Do You Evaluate Definite Integrals Of Logarithmic Functions?

You evaluate a definite log integral by finding the antiderivative and substituting the limits, exactly as for definite integrals of any other function. The one extra check is the domain: both limits must sit where the logarithm is defined.

$$\int_{1}^{e} \ln x,dx = \big[x\ln x - x\big]_{1}^{e} = (e - e) - (0 - 1) = 1$$

$$\int_{1}^{2} \ln x,dx = (2\ln 2 - 2) - (0 - 1) = 2\ln 2 - 1 \approx 0.3863$$

A Study.com question asked students to "explain why the integral from 0 to 1 of ln(x) dx is improper" and whether it converges. It is an improper integral because $\ln x \to -\infty$ as $x \to 0^{+}$, so the lower limit must be handled with a limit:

$$\int_{0}^{1} \ln x,dx = \lim_{t \to 0^{+}} \big[x\ln x - x\big]_{t}^{1}$$

$$= (0 - 1) - \lim_{t \to 0^{+}} \big(t\ln t - t\big)$$

$$= -1 - 0 = -1$$

The key fact is $\lim_{t \to 0^{+}} t\ln t = 0$: the factor $t$ shrinks to zero faster than $\ln t$ falls to $-\infty$. So the integral converges to $-1$, which is negative because the whole curve sits below the $x$-axis on $(0, 1)$.

Examples Of Integration Of Logarithmic Functions

Example 1

Find $\int \ln(5x),dx$.

Split the logarithm with the product rule for logs:

$$\ln(5x) = \ln 5 + \ln x$$

$$\int \ln(5x),dx = x\ln 5 + x\ln x - x + C$$

$$= x\ln(5x) - x + C$$

Final answer: $x\ln(5x) - x + C$

Example 2

Find $\int x\ln x,dx$.

The tempting first attempt integrates each factor separately and multiplies the results:

$$\int x,dx = \frac{x^2}{2}, \qquad \int \ln x,dx = x\ln x - x$$

$$\text{Attempt: } \frac{x^2}{2}\big(x\ln x - x\big)$$

Differentiate the attempt to test it:

$$\frac{d}{dx}\left[\frac{x^2}{2}\big(x\ln x - x\big)\right] = \frac{3x^2}{2}\ln x - x^2$$

That is not $x\ln x$, so the attempt fails. The integral of a product is not the product of the integrals, in the same way the derivative of a product is not the product of the derivatives. The rescue is integration by parts with $u = \ln x$ and $dv = x,dx$:

$$du = \frac{1}{x},dx, \qquad v = \frac{x^2}{2}$$

$$\int x\ln x,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x},dx$$

$$= \frac{x^2}{2}\ln x - \int \frac{x}{2},dx$$

$$= \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

Final answer: $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$

Example 3

Find $\int \ln(3x - 2),dx$ for $x > \frac{2}{3}$.

Substitute $u = 3x - 2$:

$$du = 3,dx, \qquad dx = \frac{du}{3}$$

$$\int \ln(3x-2),dx = \frac{1}{3}\int \ln u,du$$

$$= \frac{1}{3}\big(u\ln u - u\big) + C$$

$$= \frac{3x-2}{3}\ln(3x-2) - x + C$$

The step students skip here is the $\frac{1}{3}$ that comes from $dx = \frac{du}{3}$; without it the answer is three times too large.

Final answer: $\frac{3x-2}{3}\ln(3x-2) - x + C$

Example 4

Evaluate $\int_{1}^{4} \log_2 x,dx$.

Change the base, then use the $\ln x$ antiderivative:

$$\int_{1}^{4} \log_2 x,dx = \frac{1}{\ln 2}\big[x\ln x - x\big]_{1}^{4}$$

$$= \frac{1}{\ln 2}\Big[(4\ln 4 - 4) - (0 - 1)\Big]$$

$$= \frac{8\ln 2 - 3}{\ln 2}$$

$$= 8 - \frac{3}{\ln 2} \approx 3.6719$$

Final answer: $8 - \frac{3}{\ln 2} \approx 3.6719$

Example 5

Evaluate $\int_{1}^{e} (\ln x)^2,dx$.

Use the reduction-formula result $x(\ln x)^2 - 2x\ln x + 2x$:

$$\Big[x(\ln x)^2 - 2x\ln x + 2x\Big]_{1}^{e}$$

$$= (e - 2e + 2e) - (0 - 0 + 2)$$

$$= e - 2 \approx 0.7183$$

Final answer: $e - 2 \approx 0.7183$

Example 6

In information theory, terms of the form $x\ln x$ appear in entropy formulas. Evaluate $\int_{0}^{1} x\ln x,dx$.

The integrand is undefined at $x = 0$, so treat the lower limit with a limit and reuse the antiderivative from Example 2:

$$\int_{0}^{1} x\ln x,dx = \lim_{t \to 0^{+}} \left[\frac{x^2}{2}\ln x - \frac{x^2}{4}\right]_{t}^{1}$$

$$= \left(0 - \frac{1}{4}\right) - \lim_{t \to 0^{+}} \left(\frac{t^2}{2}\ln t - \frac{t^2}{4}\right)$$

$$= -\frac{1}{4} - 0$$

Final answer: $-\frac{1}{4}$

Why Does The Logarithm Need Its Own Integration Rule?

The short reason: the power rule breaks at exactly $n = -1$. Every power of $x$ integrates by $\int x^n,dx = \frac{x^{n+1}}{n+1} + C$, except $x^{-1}$, where the formula would divide by zero. The logarithm is what fills that hole, and once it exists, integrating it in turn needs its own tool.

• The gap in the power rule. Without the logarithm, $\int \frac{1}{x},dx$ has no answer, and neither do decay rates, growth rates or any model built on "rate proportional to amount".

• The area that behaves like a log. The area under $y = \frac{1}{x}$ from $1$ to $t$ turns products into sums, the defining property of a logarithm. That observation, made geometrically in the seventeenth century, is the historical origin of $\ln x$ as an integral.

• No reverse rule for $\ln x$ itself. Differentiation has a rule for every elementary function, but integration does not. Integrating $\ln x$ needed integration by parts, which is why the result $x\ln x - x$ looks nothing like $\ln x$.

The destination is worth seeing early. Adding up many logarithms, such as $\ln 1 + \ln 2 + \dots + \ln n$, is approximately $\int_{1}^{n} \ln x,dx = n\ln n - n + 1$, the starting point of Stirling's approximation for very large products. For $n = 10$, the exact sum is about $15.1044$ and the integral gives $14.0259$, already close.

Where Is Integration Of Logarithmic Functions Used In Real Life?

Log integrals appear wherever a quantity is measured on a logarithmic scale or built from logarithms.

• Information and entropy. Entropy formulas sum or integrate terms like $x\ln x$; the value $\int_{0}^{1} x\ln x,dx = -\frac{1}{4}$ from Example 6 is a typical building block.

• Counting very large numbers of arrangements. Approximating a sum of logarithms by $\int \ln x,dx$ turns an impossible product into a two-term formula, as in Stirling's approximation.

• Prime numbers. The logarithmic integral $\operatorname{li}(x) = \int \frac{dt}{\ln t}$ estimates how many primes lie below $x$, one of the best-known uses of a non-elementary log integral.

• Averages on log scales. The average value of a function $\ln x$ on $[1, e]$ is $\frac{1}{e-1}\int_{1}^{e} \ln x,dx = \frac{1}{e-1} \approx 0.5820$, the kind of calculation behind averaging decibel or pH readings over a range.

Where Do Students Trip Up On Integration Of Logarithmic Functions?

Mistake 1: Reversing The Derivative Instead Of Integrating

Where it slips in: Right after learning that the derivative of $\ln x$ is $\frac{1}{x}$.

Don't do this: Writing $\int \ln x,dx = \frac{1}{x} + C$, which is the derivative, not the antiderivative.

The correct way: Use parts to get $x\ln x - x + C$, then differentiate it back to confirm you return to $\ln x$.

The memorizer is most exposed here: the pair "$\ln x$ and $\frac{1}{x}$" is stored as a fact, but without its direction.

Mistake 2: Losing The Minus X During The Tidy-Up

Where it slips in: Combining terms at the end of $\int \ln(x+1),dx$, after the parts step gives $x\ln(x+1) - x + \ln(x+1) + C$.

Don't do this: Writing the tidy form as $(x+1)\ln(x+1) + C$, which silently deletes the $-x$.

The correct way: Group only the log terms: $(x+1)\ln(x+1) - x + C$. Differentiate to check: $\ln(x+1) + 1 - 1 = \ln(x+1)$.

The $-x$ vanishes most often at exactly this regrouping step, when two log terms merge and the lone $-x$ gets swept into the bracket along with them. The rusher who skips the derivative check never notices.

Mistake 3: Ignoring The Base Of The Logarithm

Where it slips in: Integrating $\log x$ or $\log_{10} x$ as if it were $\ln x$.

Don't do this: Writing $\int \log_{10} x,dx = x\log_{10} x - x + C$.

The correct way: Convert first: $\int \log_{10} x,dx = \frac{x\ln x - x}{\ln 10} + C$, equivalently $x\log_{10} x - \frac{x}{\ln 10} + C$.

The second-guesser often gets this right and then "corrects" it back to the wrong version, because the base-10 answer looks less tidy. The factor $\frac{1}{\ln 10}$ is exactly what makes it right.

Mistake 4: Plugging Zero Into X Ln X

Where it slips in: Evaluating an improper integral such as $\int_{0}^{1} x\ln x,dx$ or $\int_{0}^{1} \ln x,dx$.

Don't do this: Substituting $x = 0$ directly and declaring the answer undefined because $\ln 0$ does not exist.

The correct way: Replace the endpoint with a limit and use $\lim_{t \to 0^{+}} t\ln t = 0$.

The real-world version of Mistake 4 lives in software. Entropy calculations multiply a probability by its logarithm, and a raw computation of $0 \times \ln 0$ produces an error instead of the correct limiting value $0$. Scientific libraries handle it on purpose: the SciPy function xlogy computes $x\log y$ and returns $0$ when $x = 0$, exactly the limit rule above.

Practice Problems On Integration Of Logarithmic Functions

1. Find $\int \ln(7x),dx$.

2. Find $\int x^2\ln x,dx$.

3. Find $\int \frac{(\ln x)^3}{x},dx$.

4. Evaluate $\int_{1}^{e} x\ln x,dx$.

5. Find $\int \frac{\ln x}{\sqrt{x}},dx$.

6. Find $\int \log_{10} x,dx$.

Answer to Question 1: $x\ln(7x) - x + C$

Answer to Question 2: $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$

Answer to Question 3: $\frac{(\ln x)^4}{4} + C$

Answer to Question 4: $\frac{e^2 + 1}{4} \approx 2.0973$

Answer to Question 5: $2\sqrt{x}\ln x - 4\sqrt{x} + C$

Answer to Question 6: $\frac{x\ln x - x}{\ln 10} + C$

Key Takeaways

• Integration of logarithmic functions rests on one result: $\int \ln x,dx = x\ln x - x + C$, found by parts with $u = \ln x$ and $dv = dx$.

• Other bases convert to $\ln$ first, which adds a factor of $\frac{1}{\ln a}$.

• Logs of linear expressions use substitution, powers times $\ln x$ use parts, and any $\frac{1}{x}$ factor signals substitution.

• Improper integrals near $x = 0$ rely on the limit $t\ln t \to 0$.

• Differentiating your answer back is the fastest way to catch a lost $-x$ or a missing base factor.

Your Next Move

Work through practice problems 1 to 6; if the choice of $u$ trips you up on problem 2 or 4, return to "How Do You Integrate A Power Times The Natural Log?", and if problem 3 or 5 stalls, reread the method table in "When Does Substitution Beat Integration By Parts For Log Integrals?". Want a live Bhanzu trainer to walk through more log integrals with you? Book a free demo class.

Read More

• List of integrals for the full standard table across every function family

• Integration of exponential functions, the inverse partner of this topic

• Logarithmic differentiation for using logs to differentiate awkward products and powers

• Methods of integration to compare parts, substitution and partial fractions side by side

• Logarithmic functions for the graphs, domain and range behind every integral on this page