AP Calculus AB and BC

Integral of 1/(2x): Answer, Proof, and Steps

Q: Is (1/2)ln|x| the same as ln|x|/2?

Yes, they are identical. \frac{1}{2}\ln|x| and \frac{\ln|x|}{2} are two ways of writing the same expression.

Q: Can I write the answer as (1/2)ln|2x|?

You can, but it only adds the constant \frac{1}{2}\ln 2, which the +C already covers. The cleaner form is \frac{1}{2}\ln|x| + C.

Q: Why not use a u = 2x substitution?

You can. With u = 2x, du = 2\,dx, the integral becomes \frac{1}{2}\int\frac{1}{u}\,du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln|2x|, which simplifies to \frac{1}{2}\ln|x| up to a constant. Factoring the constant out first is quicker.

The integral of 1/(2x) is (1/2)ln|x| + C. The constant 1/2 pulls out of the integral, leaving the integral of 1/x, which is ln|x|. The absolute value is required because 1/(2x) is defined for every x except 0.

\int \frac{1}{2x}\,dx = \frac{1}{2}\ln|x| + C

Factor out the constant

The $\frac{1}{2}$ is a constant multiplier, and constants slide outside an integral untouched. What remains is the standard integral of $\frac{1}{x}$ .

\int \frac{1}{2x}\,dx = \frac{1}{2}\int \frac{1}{x}\,dx

Since $\frac{d}{dx}\ln|x| = \frac{1}{x}$ , the inside integral is $\ln|x|$ .

\int \frac{1}{2x}\,dx = \frac{1}{2}\ln|x| + C

Why the bars are not optional

The function $\frac{1}{2x}$ is defined for every $x \neq 0$ , including negatives, but $\ln x$ exists only for $x > 0$ . The absolute value extends the antiderivative to the negative side, where the chain rule on $\ln(-x)$ produces the same $\frac{1}{x}$ .

x < 0:\quad \frac{d}{dx}\ln(-x) = \frac{1}{-x}\cdot(-1) = \frac{1}{x}

Multiplying back by $\frac{1}{2}$ recovers $\frac{1}{2x}$ on both branches.

The mistake students make

The genuine error is dropping the absolute value and writing $\frac{1}{2}\ln x$ , which throws away every negative $x$ . A separate slip is folding the 2 into the log as $\frac{1}{2}\ln|2x|$ ; since $\ln|2x| = \ln 2 + \ln|x|$ , that form only adds the constant $\frac{1}{2}\ln 2$ , so it is unsimplified rather than wrong.

Constant out front, bars on the log

Pull the $\frac{1}{2}$ outside, integrate $\frac{1}{x}$ to $\ln|x|$ , and keep the bars. Only skip the absolute value when the domain is known to stay positive.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Is (1/2)ln|x| the same as ln|x|/2?

Yes, they are identical. $\frac{1}{2}\ln|x|$ and $\frac{\ln|x|}{2}$ are two ways of writing the same expression.

Can I write the answer as (1/2)ln|2x|?

You can, but it only adds the constant $\frac{1}{2}\ln 2$ , which the $+C$ already covers. The cleaner form is $\frac{1}{2}\ln|x| + C$ .

Why not use a u = 2x substitution?

You can. With $u = 2x$ , $du = 2\,dx$ , the integral becomes $\frac{1}{2}\int\frac{1}{u}\,du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln|2x|$ , which simplifies to $\frac{1}{2}\ln|x|$ up to a constant. Factoring the constant out first is quicker.