AP Calculus AB and BC

Integral of x/sqrt(1-x^2)

The integral of x over the square root of 1 minus x squared is negative the square root of 1 minus x squared, plus C. The x in the numerator is what makes this a substitution problem. Without it the answer would be arcsine instead.

∫x1−x2 dx=−1−x2+C\int \frac{x}{\sqrt{1-x^{2}}}\,dx = -\sqrt{1-x^{2}} + C

The substitution

u=1−x2,du=−2x dx  ⟹  x dx=−du2u = 1-x^{2}, \quad du = -2x\,dx \implies x\,dx = -\frac{du}{2}
∫x dx1−x2=−12∫u−1/2du=−u1/2+C=−1−x2+C\int \frac{x\,dx}{\sqrt{1-x^{2}}} = -\frac{1}{2}\int u^{-1/2}du = -u^{1/2} + C = -\sqrt{1-x^{2}} + C

One character changes everything

Remove the xx and the problem becomes a completely different one, with an inverse trigonometric answer rather than an algebraic one.

∫dx1−x2=arcsin⁡x+C,∫x dx1−x2=−1−x2+C\int \frac{dx}{\sqrt{1-x^{2}}} = \arcsin x + C, \qquad \int \frac{x\,dx}{\sqrt{1-x^{2}}} = -\sqrt{1-x^{2}} + C

Check the numerator first

Before reaching for an arcsine pattern, look at what is on top. A bare constant gives the inverse trig form; a matching x makes it an ordinary substitution.

Common mistakes

  • Answering arcsin⁡x\arcsin x by pattern-matching the denominator alone.
  • Losing the minus sign from du=−2x dxdu = -2x\,dx.
  • Forgetting the domain: the integrand needs −1<x<1-1 < x < 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(1-x^2)?

It is −1−x2+C-\sqrt{1-x^{2}} + C.

Why is it not arcsin x?

That is the answer when the numerator is 11. The extra xx makes it an ordinary substitution instead.