AP Calculus AB and BC

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

Q: Why is there no absolute value in the answer?

Because x^2 + 1 is positive for every real x, so |x^2 + 1| = x^2 + 1. The bars in \ln|u| only matter when the argument can turn negative.

Q: How is x/(x^2+1) different from 1/(x^2+1)?

The x on top changes the technique. With x, use u = x^2 + 1 to get \frac{1}{2}\ln(x^2+1) + C. Without it, \int \frac{1}{x^2+1}\,dx = \arctan x + C.

The integral of x/(x^2+1) is (1/2)ln(x^2+1) + C. Substituting u = x^2+1 gives du = 2x dx, so x dx = du/2 and the integral becomes one half times the integral of 1/u. No absolute value is needed because x^2+1 is always positive. Differentiating (1/2)ln(x^2+1) returns x/(x^2+1).

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

The substitution, step by step

The numerator $x$ is, up to a constant, the derivative of the denominator $x^2 + 1$ . That match is the signal for a u-substitution.

u = x^2 + 1 \quad\Longrightarrow\quad \frac{du}{dx} = 2x \quad\Longrightarrow\quad x\,dx = \frac{du}{2}

Replace $x^2 + 1$ with $u$ and $x\,dx$ with $\frac{du}{2}$ , leaving the basic logarithm integral.

\int \frac{x}{x^2 + 1}\,dx = \int \frac{1}{u}\cdot\frac{du}{2} = \frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln|u| + C

Substituting $u = x^2 + 1$ back gives the antiderivative.

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

A derivative on top means a log

When the numerator is the derivative of the denominator, the integral is a natural log of that denominator. The $\frac{1}{2}$ comes from the constant in $du = 2x\,dx$ .

Why no absolute value is needed

The logarithm rule normally carries absolute value bars, as in $\int \frac{1}{u}\,du = \ln|u| + C$ . Here the substitution variable is always positive.

u = x^2 + 1 \geq 1 > 0 \quad \text{for every real } x

Since $x^2 + 1$ can never be zero or negative, $|x^2 + 1| = x^2 + 1$ , and the bars drop away safely. The antiderivative $\frac{1}{2}\ln(x^2 + 1)$ is defined for all real $x$ .

Common mistakes

Forgetting the $\frac{1}{2}$ . The constant from $du = 2x\,dx$ produces it; without it the derivative comes out as $\frac{2x}{x^2+1}$ .
Keeping unnecessary absolute value bars. Because $x^2 + 1 > 0$ always, $\ln(x^2+1)$ is correct without them.
Splitting off $\frac{1}{x^2+1}$ and reaching for arctangent. The $x$ on top makes this a logarithm substitution, not the arctangent form.

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

What is the integral of x/(x^2+1) from 0 to 1?

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

Why is there no absolute value in the answer?

Because $x^2 + 1$ is positive for every real $x$ , so $|x^2 + 1| = x^2 + 1$ . The bars in $\ln|u|$ only matter when the argument can turn negative.

How is x/(x^2+1) different from 1/(x^2+1)?

The $x$ on top changes the technique. With $x$ , use $u = x^2 + 1$ to get $\frac{1}{2}\ln(x^2+1) + C$ . Without it, $\int \frac{1}{x^2+1}\,dx = \arctan x + C$ .