AP Calculus AB and BC

Integral of 1/(x+1): Answer, Proof, and Mistakes

Q: Why does the integral of 1/(x+1) need an absolute value?

Because \frac{1}{x+1} is defined for x < -1, where x+1 is negative, and \ln(x+1) is not. The bars extend the antiderivative onto that side, since \frac{d}{dx}\ln|x+1| = \frac{1}{x+1} everywhere except x = -1.

Q: What is the integral of 1/(x+1) from 0 to 1?

\int_0^1 \frac{1}{x+1}\,dx = \left[\ln|x+1|\right]_0^1 = \ln 2 - \ln 1 = \ln 2.

Q: What is the integral of 1/(ax+b)?

\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C. The substitution u = ax+b gives du = a\,dx, producing the \frac{1}{a}. When a = 1, as in \frac{1}{x+1}, no factor appears.

The integral of 1/(x+1) is ln|x+1| + C. The log rule for a linear denominator gives the natural log of the absolute value, and the absolute value is required because x+1 can be negative. Differentiating ln|x+1| gives 1/(x+1).

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

Applying the log rule

The reciprocal of a linear expression integrates to a natural log. The base rule is $\int \frac{1}{x}\,dx = \ln|x| + C$ , and a substitution handles the shift by $1$ .

u = x + 1 \quad\Longrightarrow\quad du = dx

Because $\frac{du}{dx} = 1$ , the differential passes straight through with no extra factor. The integral becomes the base log rule in $u$ .

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

No reciprocal here

The inner coefficient is $1$ , so no factor appears. In general $\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C$ .

Why the absolute value is required

The function $\frac{1}{x+1}$ is defined for every $x \neq -1$ , including where $x + 1$ is negative. Its antiderivative has to live on that same two-sided domain, and only $\ln|x+1|$ does.

\frac{d}{dx}\left[\ln|x+1|\right] = \frac{1}{x+1}, \quad x \neq -1

For $x + 1 > 0$ the bars do nothing. For $x + 1 < 0$ , the chain rule on $\ln(-(x+1))$ contributes a $-1$ that cancels the negative inside, so the derivative is still $\frac{1}{x+1}$ .

The bars are not decoration

Writing $\ln(x+1)$ without bars silently discards every $x < -1$ , where $x+1$ is negative and the plain log is undefined.

Common mistakes

Dropping the absolute value. $\ln(x+1)$ is undefined for $x < -1$ , but $\frac{1}{x+1}$ is not, so the bars are mandatory.
Writing $\ln|x| + 1$ or similar. The whole denominator $x+1$ goes inside the log, giving $\ln|x+1|$ .
Integrating across $x = -1$ . The integrand has a vertical asymptote there, so a definite integral spanning $-1$ is improper and generally diverges.

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

Why does the integral of 1/(x+1) need an absolute value?

Because $\frac{1}{x+1}$ is defined for $x < -1$ , where $x+1$ is negative, and $\ln(x+1)$ is not. The bars extend the antiderivative onto that side, since $\frac{d}{dx}\ln|x+1| = \frac{1}{x+1}$ everywhere except $x = -1$ .

What is the integral of 1/(x+1) from 0 to 1?

$\int_0^1 \frac{1}{x+1}\,dx = \left[\ln|x+1|\right]_0^1 = \ln 2 - \ln 1 = \ln 2$ .

What is the integral of 1/(ax+b)?

$\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C$ . The substitution $u = ax+b$ gives $du = a\,dx$ , producing the $\frac{1}{a}$ . When $a = 1$ , as in $\frac{1}{x+1}$ , no factor appears.