AP Calculus AB and BC

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

The integral of x sqrt(x^2+1) is (x^2+1)^(3/2)/3 + C. Let u = x^2 + 1, so du = 2x dx and the x dx in the integrand becomes du/2. The integral turns into one half the integral of the square root of u du, which gives u^(3/2)/3.

∫xx2+1 dx=(x2+1)3/23+C\int x\sqrt{x^2+1}\,dx = \frac{(x^2+1)^{3/2}}{3} + C

Substituting u = x^2 + 1

The expression under the root is x2+1x^2+1, and its derivative is 2x2x. The integrand already carries an xx, which is 2x2x up to a constant, so substitution clears the whole integral.

u=x2+1,du=2x dx,x dx=du2u = x^2+1, \qquad du = 2x\,dx, \qquad x\,dx = \frac{du}{2}

Replace x2+1\sqrt{x^2+1} with u\sqrt{u} and x dxx\,dx with du2\frac{du}{2}, then integrate u1/2u^{1/2} with the power rule.

∫xx2+1 dx=12∫u1/2 du=12⋅23u3/2+C=(x2+1)3/23+C\int x\sqrt{x^2+1}\,dx = \frac{1}{2}\int u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3}u^{3/2} + C = \frac{(x^2+1)^{3/2}}{3} + C

Where the 3 comes from

Two constants meet and simplify: the 12\frac{1}{2} from x dx=du2x\,dx = \frac{du}{2} and the 23\frac{2}{3} from raising u1/2u^{1/2} to u3/2u^{3/2}. Their product is 13\frac{1}{3}.

Checking by differentiating

Apply the chain rule to the result and confirm the integrand returns.

ddx((x2+1)3/23)=13⋅32(x2+1)1/2⋅2x=xx2+1\frac{d}{dx}\left(\frac{(x^2+1)^{3/2}}{3}\right) = \frac{1}{3}\cdot\frac{3}{2}(x^2+1)^{1/2}\cdot 2x = x\sqrt{x^2+1}

The 32\frac{3}{2} from the power rule and the inner derivative 2x2x combine with the 13\frac{1}{3} out front to leave a single factor of xx.

The mistakes students make

  • Forgetting the 12\frac{1}{2} from x dx=du2x\,dx = \frac{du}{2}, which gives 2(x2+1)3/23\frac{2(x^2+1)^{3/2}}{3}, twice the correct antiderivative.
  • Integrating the root and the xx separately, as if ∫fg=∫f∫g\int fg = \int f\int g. No such rule exists.
  • Treating x2+1\sqrt{x^2+1} as x+1x+1. The root of a sum is not the sum of the roots, and that slip quietly replaces the problem with ∫(x2+x) dx\int (x^2+x)\,dx.
  • Jumping to trigonometric substitution with x=tan⁡θx = \tan\theta. It works, but the xx sitting outside the root makes plain substitution much faster.

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 substitution work here but not for sqrt(x^2+1) alone?

Substitution needs the inner function's derivative present. Here x dxx\,dx supplies du2\frac{du}{2}. Without the outside xx there is nothing to absorb dudu, and ∫x2+1 dx\int\sqrt{x^2+1}\,dx requires trigonometric substitution instead.

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

∫02xx2+1 dx=[(x2+1)3/23]02=53/2−13=55−13\int_0^2 x\sqrt{x^2+1}\,dx = \left[\frac{(x^2+1)^{3/2}}{3}\right]_0^2 = \frac{5^{3/2} - 1}{3} = \frac{5\sqrt{5} - 1}{3}.

Can the answer be written a different way?

Yes. (x2+1)3/23\frac{(x^2+1)^{3/2}}{3}, 13(x2+1)x2+1\frac{1}{3}(x^2+1)\sqrt{x^2+1}, and 13(x2+1)3\frac{1}{3}\sqrt{(x^2+1)^3} are the same function, so any of them earns full credit.