AP Calculus AB and BC

Integral of 1/sqrt(x+1): Answer and Rule

Q: Why the factor of 2?

The new exponent is \frac{1}{2}, and dividing by \frac{1}{2} multiplies by 2.

The integral of 1 over the square root of x plus 1 is 2 times the square root of x plus 1, plus C. Rewriting the integrand as x plus 1 to the power negative one half lets the power rule apply, and the shift costs nothing because its derivative is 1.

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

Rewrite as a power

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

Dividing by $\frac{1}{2}$ is multiplying by $2$ , which is where the leading coefficient comes from.

An improper integral worth knowing

$\int_{0}^{\infty}\frac{dx}{\sqrt{x+1}}$ diverges, because $2\sqrt{x+1}$ grows without bound. Compare $\frac{1}{(x+1)^{2}}$ , which converges: the exponent decides, exactly as with a $p$ -series.

Common mistakes

Answering $\frac{1}{2}\sqrt{x+1}$ , dividing instead of multiplying.
Producing a logarithm. That only happens for the exponent $-1$ .

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 1/sqrt(x+1)?

It is $2\sqrt{x+1} + C$ .

Why the factor of 2?

The new exponent is $\frac{1}{2}$ , and dividing by $\frac{1}{2}$ multiplies by $2$ .