AP Calculus AB and BC

Integral of sqrt(x+1): The Free Substitution

Q: What is the integral of sqrt(x+1)?

It is \frac{2}{3}\left(x+1\right)^{3/2} + C.

Q: Do I actually need u substitution for sqrt(x+1)?

Formally yes, in practice no. Since du = dx, the reverse power rule applies directly to x+1. Writing the substitution out is still worth doing when the inside is anything more complicated.

Q: Is sqrt(x+1) the same as sqrt(x) + 1?

No. Test x = 9: \sqrt{10} is about 3.162, while \sqrt{9}+1 = 4. The two functions differ everywhere except at x = 0.

The integral of the square root of x plus one is two thirds of the quantity x plus one raised to the three halves power, plus C. Substituting u equal to x plus one gives du equal to dx, so nothing is left over to divide out and the reverse power rule finishes the job.

\int \sqrt{x+1}\,dx = \frac{2}{3}\left(x+1\right)^{3/2} + C

u = x + 1, and du = dx

Every substitution has a bookkeeping cost: the $dx$ has to be converted into $du$ . Here that cost is nothing at all. With $u = x+1$ , $\frac{du}{dx} = 1$ , so $du = dx$ and the integral becomes a bare power of $u$ .

\int\sqrt{x+1}\,dx = \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}\left(x+1\right)^{3/2} + C

Because the inside is $x+1$ and not something like $3x+1$ , no constant has to be divided out at the end. In $\int\sqrt{3x+1}\,dx$ the relation $du = 3\,dx$ forces an extra factor of $\frac{1}{3}$ , which is exactly the step people forget.

Why the answer is not two thirds x to the three halves plus x

The square root does not distribute over addition. $\sqrt{x+1}$ is not $\sqrt{x}+1$ : at $x = 9$ the first is $\sqrt{10}\approx 3.162$ and the second is $4$ . Any antiderivative built by splitting the root is answering a different question.

So the answer is not $\frac{2}{3}x^{3/2} + x + C$ either. The whole quantity $x+1$ is what gets raised to the $\frac{3}{2}$ power, and it stays glued together from the first line to the last.

The thirty second check

Differentiate the answer. Two thirds times three halves is one, the power drops to one half, and the inner derivative is 1, so you get the square root of x plus one back. Any missing constant or split root shows up immediately.

The mistakes students make

This integrand is easy enough that errors come from rushing rather than from confusion about method.

Splitting the root and answering $\frac{2}{3}x^{3/2} + x + C$ , which is the antiderivative of $\sqrt{x}+1$ .
Raising the power correctly but forgetting to divide, giving $\left(x+1\right)^{3/2} + C$ , whose derivative is $\frac{3}{2}\sqrt{x+1}$ .
Ignoring the domain. The integrand is undefined for $x < -1$ , so a definite integral from $-3$ to $0$ has no value.

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 sqrt(x+1)?

It is $\frac{2}{3}\left(x+1\right)^{3/2} + C$ .

Do I actually need u substitution for sqrt(x+1)?

Formally yes, in practice no. Since $du = dx$ , the reverse power rule applies directly to $x+1$ . Writing the substitution out is still worth doing when the inside is anything more complicated.

Is sqrt(x+1) the same as sqrt(x) + 1?

No. Test $x = 9$ : $\sqrt{10}$ is about $3.162$ , while $\sqrt{9}+1 = 4$ . The two functions differ everywhere except at $x = 0$ .