List of Integrals of Logarithmic Functions With Proofs

What Is The List Of Integrals Of Logarithmic Functions?

The list of integrals of logarithmic functions is a reference table of antiderivatives for integrands that contain a logarithm, such as $\ln x$, $\log_a x$, $(\ln x)^n$, $x^m\ln x$ and $\frac{\ln x}{x}$. Its anchor entry is

$$\int \ln x,dx = x\ln x - x + C, \qquad x > 0,$$

where $\ln x$ is the natural logarithm and $C$ is the constant of integration. Every other entry is this result after a change of base, a substitution or one more round of integration by parts.

The symbols used throughout the tables:

How Do You Read A Table Of Logarithmic Integrals?

You read each row as "the derivative of the right-hand column is the integrand", valid only inside the stated domain. Three reading rules prevent most errors.

• The domain column is part of the answer. $\ln x$ needs $x > 0$, so a row like $\int \ln x,dx$ says nothing about negative $x$. Rows that output $\ln\lvert u\rvert$ keep the absolute value because the integrand is defined on both sides of zero.

• "log" means $\ln$ in this list. University texts and software usually write $\log$ for base $e$; for base 10 or any base $a$, use the change-of-base rows.

• Different-looking answers can both be right. Two antiderivatives of the same function may differ by a constant, so $x\ln(2x) - x$ and $x\ln x - x + x\ln 2$ are the same entry.

That last rule answers a Quora question that comes up often: "Why is the integral of 1/x ln(x) instead of ln(ax)?" Both are correct. Since $\ln(ax) = \ln x + \ln a$ and $\ln a$ is a constant, $\ln(ax)$ and $\ln x$ differ only by a constant that $C$ absorbs.

What Are The Integrals Involving Only Logarithms?

These rows integrate a single logarithm of $x$, of a linear expression or of a simple quadratic. All of them come from integration by parts with $u$ equal to the logarithm and $dv = dx$, sometimes after a substitution.

The change-of-base step uses one of the properties of logarithms, $\log_a x = \frac{\ln x}{\ln a}$. For base 10 the factor $\frac{1}{\ln 10}$ is about $0.4343$.

The $\ln(x^2+a^2)$ row deserves a second look because it is the one most likely to be misremembered. Parts gives $x\ln(x^2+a^2) - \int \frac{2x^2}{x^2+a^2},dx$, and rewriting $\frac{2x^2}{x^2+a^2} = 2 - \frac{2a^2}{x^2+a^2}$ produces both the $-2x$ and the $2a\arctan\frac{x}{a}$.

What Are The Integrals Of Powers Of Logarithms?

Powers of $\ln x$ are integrated with a reduction formula, an identity that lowers the power by one each time it is applied:

$$\int (\ln x)^n,dx = x(\ln x)^n - n\int (\ln x)^{n-1},dx, \qquad x > 0.$$

It comes from parts with $u = (\ln x)^n$ and $dv = dx$, since $du = \frac{n(\ln x)^{n-1}}{x},dx$ and $v = x$ cancel the $\frac{1}{x}$.

The coefficients follow a falling-product pattern: for $n = 4$ they are $1$, $4$, $4 \times 3$, $4 \times 3 \times 2$ and $4 \times 3 \times 2 \times 1$, with alternating signs. That pattern is a fast check on any long answer.

What Are The Integrals Of Logarithms Times Powers Of X?

When a power of $x$ multiplies $\ln x$, parts with $u = \ln x$ and $dv = x^m,dx$ gives one general row, valid for $m \neq -1$ and $x > 0$:

$$\int x^m\ln x,dx = \frac{x^{m+1}}{m+1}\ln x - \frac{x^{m+1}}{(m+1)^2} + C.$$

The excluded value $m = -1$ is not a gap in the list. It is the integrand $\frac{\ln x}{x}$, which belongs to the substitution family below.

Which Logarithmic Integrals Use Substitution?

Any integrand that pairs a function of $\ln x$ with a factor $\frac{1}{x}$ uses integration by substitution with $u = \ln x$, because $du = \frac{1}{x},dx$. A Quora question, "How to integrate 1/x ln(x)", is the classic case: with $u = \ln x$ it becomes $\int \frac{du}{u} = \ln\lvert u\rvert + C$.

Two homework questions in the search results show how far this family reaches. The first asks for $\int \frac{\ln(x+1) - \ln x}{x(x+1)},dx$. Set $t = \ln\left(1 + \frac{1}{x}\right)$, which equals the numerator, and note that $dt = -\frac{1}{x(x+1)},dx$:

$$\int \frac{\ln(x+1) - \ln x}{x(x+1)},dx = -\int t,dt = -\frac{1}{2}\left[\ln\left(1 + \frac{1}{x}\right)\right]^2 + C.$$

The second asks for $\int \frac{\ln x}{(1+\ln x)^2},dx$. No substitution is needed once you spot that it is a quotient-rule derivative: $\frac{d}{dx}\left[\frac{x}{1+\ln x}\right] = \frac{(1+\ln x) - 1}{(1+\ln x)^2} = \frac{\ln x}{(1+\ln x)^2}$, so the answer is $\frac{x}{1+\ln x} + C$.

What Are The Integrals Of Logarithms With Trig, Exponential And Hyperbolic Functions?

These mixed rows either substitute $x = e^{t}$ or recognise a product-rule derivative. Each one is easiest to trust by differentiating the result.

The hyperbolic functions rows look exotic but collapse quickly: $\sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2} = \frac{1}{2}\left(x - \frac{1}{x}\right)$, which integrates term by term. The two exponential rows are the product rule read backwards, since $\frac{d}{dx}\big[e^{x}g(x)\big] = e^{x}\big(g(x) + g'(x)\big)$.

Which Logarithmic Integrals Have No Elementary Formula?

Some logarithmic integrands have antiderivatives that cannot be written with a finite combination of polynomials, roots, exponentials, logarithms and trig functions. A Quora thread asks it directly: "Is it possible to integrate 1/ln(x)?" It is, but the answer is a new function rather than a formula built from old ones.

The logarithmic integral is defined as a principal-value integral, $\operatorname{li}(x) = \int_{0}^{x} \frac{dt}{\ln t}$, in the NIST Digital Library of Mathematical Functions, equation 6.2.8. Numerical values: $\operatorname{li}(2) \approx 1.0452$, $\operatorname{li}(10) \approx 6.1656$, $\operatorname{li}(100) \approx 30.1261$. Its only positive zero, the Ramanujan-Soldner constant, is about $1.4514$.

What Are The Standard Definite Integrals Of Logarithms?

Many logarithmic integrals over $[0, 1]$ are improper integrals, because $\ln x \to -\infty$ as $x \to 0^{+}$. They still converge, thanks to the limit $\lim_{t \to 0^{+}} t^{k}\ln t = 0$ for any $k > 0$.

The pattern in the first three rows continues: $\int_{0}^{1} (\ln x)^n,dx$ equals $n$ factorial with the sign $(-1)^n$, so $n = 4$ gives $24$. The rows with $\pi^2$ come from expanding the logarithm as a power series and summing term by term, an idea that belongs to series courses rather than to this list. For more on evaluating with limits, see definite integrals.

Examples Of Integrals Of Logarithmic Functions

Example 1

Use the list to find $\int \ln(4x + 1),dx$.

Match the row $\int \ln(ax+b),dx$ with $a = 4$ and $b = 1$:

$$\int \ln(4x+1),dx = \frac{4x+1}{4}\ln(4x+1) - x + C$$

The domain is $4x + 1 > 0$, so $x > -\frac{1}{4}$.

Final answer: $\frac{4x+1}{4}\ln(4x+1) - x + C$

Example 2

Find $\int \frac{dx}{x\ln x}$ for $x > 1$.

A quick scan of the table can land on the wrong row. The integrand contains $\frac{1}{\ln x}$, and the row $\int \frac{dx}{\ln x} = \operatorname{li}(x) + C$ says that has no elementary formula, so a hurried reader concludes:

$$\text{Attempt: } \int \frac{dx}{x\ln x} \text{ has no elementary antiderivative.}$$

That conclusion ignores the extra $\frac{1}{x}$. The extra factor changes the family entirely, because $\frac{1}{x},dx$ is exactly $d(\ln x)$. Substitute $u = \ln x$:

$$du = \frac{1}{x},dx$$

$$\int \frac{dx}{x\ln x} = \int \frac{du}{u}$$

$$= \ln\lvert u\rvert + C = \ln\lvert\ln x\rvert + C$$

Check: $\frac{d}{dx}\ln(\ln x) = \frac{1}{\ln x}\cdot\frac{1}{x}$, which is the integrand.

Final answer: $\ln\lvert\ln x\rvert + C$

Example 3

Use the list to find $\int x^3\ln x,dx$.

Match the row $\int x^m\ln x,dx$ with $m = 3$:

$$\int x^3\ln x,dx = \frac{x^4}{4}\ln x - \frac{x^4}{4^2} + C$$

$$= \frac{x^4}{4}\ln x - \frac{x^4}{16} + C$$

Final answer: $\frac{x^4}{4}\ln x - \frac{x^4}{16} + C$

Example 4

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

Use the row $\int \frac{(\ln x)^n}{x},dx = \frac{(\ln x)^{n+1}}{n+1} + C$ with $n = 2$:

$$\int_{1}^{e} \frac{(\ln x)^2}{x},dx = \left[\frac{(\ln x)^3}{3}\right]_{1}^{e}$$

$$= \frac{1}{3} - 0 = \frac{1}{3}$$

Final answer: $\frac{1}{3} \approx 0.3333$

Example 5

Evaluate $\int_{0}^{1} x^2\ln x,dx$.

Use the definite row $\int_{0}^{1} x^m\ln x,dx = -\frac{1}{(m+1)^2}$ with $m = 2$, or work from the indefinite row:

$$\int_{0}^{1} x^2\ln x,dx = \lim_{t \to 0^{+}} \left[\frac{x^3}{3}\ln x - \frac{x^3}{9}\right]_{t}^{1}$$

$$= \left(0 - \frac{1}{9}\right) - 0 = -\frac{1}{9}$$

Final answer: $-\frac{1}{9} \approx -0.1111$

Example 6

A designer models a cross-section whose height is $\ln(x^2 + 9)$ for $0 \leq x \leq 3$. Find the area under the curve.

Use the row $\int \ln(x^2+a^2),dx$ with $a = 3$:

$$\int_{0}^{3} \ln(x^2+9),dx = \left[x\ln(x^2+9) - 2x + 6\arctan\frac{x}{3}\right]_{0}^{3}$$

$$= 3\ln 18 - 6 + 6\cdot\frac{\pi}{4}$$

$$= 3\ln 18 - 6 + \frac{3\pi}{2} \approx 7.3835$$

Final answer: $3\ln 18 - 6 + \frac{3\pi}{2} \approx 7.3835$ square units

Why Does A Separate List For Logarithmic Integrals Exist?

Logarithmic integrals get their own list because the logarithm has no reverse power rule. Polynomials, exponentials and trig functions each integrate by reversing one derivative rule, but $\ln x$ does not, so every log entry is the product of a technique rather than a lookup of a reversed derivative.

• One technique per family. Parts handles bare logs and powers of $x$ times logs; substitution handles anything with a $\frac{1}{x}$ factor; the reduction formula handles powers of the log.

• A boundary you can see. The list shows exactly where closed forms stop. $\int \frac{dx}{x\ln x}$ is elementary while $\int \frac{dx}{\ln x}$ is not, and the only difference is one factor of $\frac{1}{x}$.

• A destination beyond the classroom. The non-elementary entry $\operatorname{li}(x)$ is the main term in the prime number estimate: it approximates how many primes lie below $x$, giving about $177.6$ at $x = 1000$, where the true count is $168$.

Common Errors When Working With Integrals Of Logarithmic Functions

Mistake 1: Using The Power-Times-Log Row At M Equals Minus One

Where it slips in: Integrating $\frac{\ln x}{x}$ by plugging $m = -1$ into the $x^m\ln x$ row.

Don't do this: Writing $\frac{x^{0}}{0}\ln x - \frac{x^{0}}{0^2}$, which divides by zero.

The correct way: Switch families: with $u = \ln x$, $\int \frac{\ln x}{x},dx = \frac{(\ln x)^2}{2} + C$.

The memorizer reaches for the general row because it "covers every power", and the condition $m \neq -1$ is the one line they skipped.

Mistake 2: Dropping The Absolute Value In Ln Of Ln X

Where it slips in: Answering $\int \frac{dx}{x\ln x}$ on an interval where $0 < x < 1$.

Don't do this: Writing $\ln(\ln x) + C$ there, which is undefined because $\ln x < 0$.

The correct way: Keep $\ln\lvert\ln x\rvert + C$, valid on $(0, 1)$ and on $(1, \infty)$ separately.

Mistake 3: Rejecting A Correct Answer Because It Looks Different

Where it slips in: Comparing your answer for $\int \ln(2x),dx$ with a table that writes it another way.

Don't do this: Deciding $x\ln x + x\ln 2 - x$ is wrong because the table shows $x\ln(2x) - x$.

The correct way: Subtract the two candidates; if the difference is a constant (here it is $0$), both are valid.

The second-guesser is the student most likely to erase a correct answer here. Differentiating both forms takes ten seconds and settles it, which is the habit worth building before any table lookup.

Mistake 4: Trusting A Pattern Without Proof

Where it slips in: Reading the $\operatorname{li}(x)$ row and assuming, from a few computed values, that $\operatorname{li}(x)$ always exceeds the number of primes below $x$.

Don't do this: Treating numerical agreement on small inputs as a theorem.

The correct way: State only what is proved: $\operatorname{li}(x)$ is asymptotically close to the prime count, and the sign of the difference changes infinitely often.

The real-world version of Mistake 4 is one of the most famous in number theory. Every computed value of $\operatorname{li}(x)$ for the prime count sat above the true count, and many expected that to hold forever, until Littlewood's 1914 proof showed it fails. The first crossing is so far out that bounding it took decades of work, described in the history of Skewes's number.

Practice Problems On The List Of Integrals Of Logarithmic Functions

1. Find $\int \ln(2x - 5),dx$ for $x > \frac{5}{2}$.

2. Find $\int (\ln x)^3,dx$.

3. Find $\int \frac{dx}{x(\ln x)^2}$ for $x > 1$.

4. Evaluate $\int_{0}^{1} (\ln x)^4,dx$.

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

Answer to Question 1: $\frac{2x-5}{2}\ln(2x-5) - x + C$

Answer to Question 2: $x\big[(\ln x)^3 - 3(\ln x)^2 + 6\ln x - 6\big] + C$

Answer to Question 3: $-\frac{1}{\ln x} + C$

Answer to Question 4: $24$

Answer to Question 5: $\frac{x}{2}\big[\sin(\ln x) + \cos(\ln x)\big] + C$

Bottom Line

• The list of integrals of logarithmic functions grows from one result, $\int \ln x,dx = x\ln x - x + C$.

• Other bases divide by $\ln a$, linear insides divide by $a$, and powers of the log use the reduction formula.

• A spare $\frac{1}{x}$ factor signals substitution with $u = \ln x$; without it, reach for integration by parts.

• $\int \frac{dx}{\ln x}$ has no elementary formula and defines $\operatorname{li}(x)$.

• Domain conditions and absolute values are part of every entry, not decoration.

Where To Go From Here

Work through practice problems 1 to 5; if you cannot tell which row a problem matches, return to "Which Logarithmic Integrals Use Substitution?" and look for the $\frac{1}{x}$ factor before anything else. Want a live Bhanzu trainer to help you build table-reading speed? Book a free demo class.

Read More

• List of integrals for every function family in one table

• Table of integrals for the standard trig, inverse-trig and quadratic forms

• Basic integration formulas for the power, exponential and reciprocal rules the log rows build on

• Integration of exponential functions, the inverse partner of the logarithm

• Derivatives of logarithmic functions for checking any entry by differentiating back