AP Calculus AB and BC

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

Q: Why is the integral from -a to a equal to zero?

The integrand is odd, since replacing x with -x flips its sign, and the antiderivative \sqrt{x^2+1} is even. Evaluating gives \sqrt{a^2+1} - \sqrt{a^2+1} = 0 for any a.

Q: Does the answer need absolute value bars?

No. The quantity x^2+1 is positive for every real x, so \sqrt{x^2+1} is always defined and the antiderivative is valid on the whole real line.

The integral of x/sqrt(x^2+1) is sqrt(x^2+1) + C. Let u = x^2 + 1, so du = 2x dx and x dx = du/2, turning the problem into one half the integral of u to the negative one half du, which equals the square root of u. Differentiating sqrt(x^2+1) gives x/sqrt(x^2+1).

\int \frac{x}{\sqrt{x^2+1}}\,dx = \sqrt{x^2+1} + C

Substituting u = x^2 + 1

The derivative of $x^2+1$ is $2x$ , and an $x$ already sits in the numerator. That is the setup substitution needs.

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

Write the root as a power so the power rule applies directly.

\int \frac{x}{\sqrt{x^2+1}}\,dx = \frac{1}{2}\int u^{-1/2}\,du = \frac{1}{2}\cdot 2u^{1/2} + C = \sqrt{x^2+1} + C

The coefficients cancel

Integrating $u^{-1/2}$ produces $2u^{1/2}$ , and that $2$ cancels the $\frac{1}{2}$ from $x\,dx = \frac{du}{2}$ . The answer is the bare square root.

Checking by differentiating

Differentiate $\sqrt{x^2+1}$ with the chain rule.

\frac{d}{dx}\sqrt{x^2+1} = \frac{1}{2}(x^2+1)^{-1/2}\cdot 2x = \frac{x}{\sqrt{x^2+1}}

The integrand comes back exactly, so $\sqrt{x^2+1}$ is the antiderivative and no coefficient is missing.

The mistakes students make

Forgetting that $\int u^{-1/2}\,du = 2u^{1/2}$ and answering $\frac{1}{2}\sqrt{x^2+1}$ , which is half the correct antiderivative and differentiates to half the integrand.
Answering $2\sqrt{x^2+1}$ by using the $2$ but skipping the $\frac{1}{2}$ from $x\,dx = \frac{du}{2}$ .
Confusing this with $\int\frac{1}{\sqrt{x^2+1}}\,dx$ , which has no $x$ on top and is a much harder integral requiring trigonometric substitution.
Answering $\ln\left|\sqrt{x^2+1}\right|$ by reaching for a logarithm. Logarithms come from $\int\frac{du}{u}$ , and the root here makes the power $-\frac{1}{2}$ , not $-1$ .

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/sqrt(x^2+1) from 0 to 1?

$\int_0^1 \frac{x}{\sqrt{x^2+1}}\,dx = \left[\sqrt{x^2+1}\right]_0^1 = \sqrt{2} - 1 \approx 0.414$ .

Why is the integral from -a to a equal to zero?

The integrand is odd, since replacing $x$ with $-x$ flips its sign, and the antiderivative $\sqrt{x^2+1}$ is even. Evaluating gives $\sqrt{a^2+1} - \sqrt{a^2+1} = 0$ for any $a$ .

Does the answer need absolute value bars?

No. The quantity $x^2+1$ is positive for every real $x$ , so $\sqrt{x^2+1}$ is always defined and the antiderivative is valid on the whole real line.