AP Calculus AB and BC

Integral of x sqrt(x+1): Substitution

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

It is \frac{2(x+1)^{5/2}}{5} - \frac{2(x+1)^{3/2}}{3} + C.

Q: What do I do with the leftover x?

Rewrite it in terms of u: since u = x+1, the stray x is u - 1.

The integral of x times the square root of x plus 1 is 2 times x plus 1 to the five halves over 5, minus 2 times x plus 1 to the three halves over 3, plus C. Substituting u equal to x plus 1 leaves a stray x, which you rewrite as u minus 1.

\int x\sqrt{x+1}\,dx = \frac{2(x+1)^{5/2}}{5} - \frac{2(x+1)^{3/2}}{3} + C

Convert the leftover factor too

With $u = x+1$ the radical becomes $\sqrt{u}$ , but the stray $x$ outside has to be converted as well: $x = u - 1$ .

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

= \frac{2u^{5/2}}{5} - \frac{2u^{3/2}}{3} + C

The step people skip

A substitution is only complete when EVERY $x$ is gone. Leaving the outside $x$ in place produces a mixed expression that cannot be integrated, and it is the single most common failure in this family of problems.

Common mistakes

Leaving the stray $x$ unconverted.
Expanding $(u-1)u^{1/2}$ incorrectly; the exponents add, giving $u^{3/2}$ .

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+1)?

It is $\frac{2(x+1)^{5/2}}{5} - \frac{2(x+1)^{3/2}}{3} + C$ .

What do I do with the leftover x?

Rewrite it in terms of $u$ : since $u = x+1$ , the stray $x$ is $u - 1$ .