AP Calculus AB and BC

Integral of x^(3/2): Answer, Proof, and Steps

Q: How do I integrate x times the square root of x?

Rewrite it first: x\sqrt{x} = x\cdot x^{1/2} = x^{3/2}, so the answer is \frac{2}{5}x^{5/2} + C. Radicals almost always become fractional exponents before any integration rule applies.

Q: Does the power rule ever fail for fractional exponents?

No. The only exponent it cannot handle is n = -1, where n+1 = 0 and the division is impossible. That single case gives \int x^{-1}\,dx = \ln|x| + C.

The integral of x^(3/2) is (2/5)x^(5/2) + C. The reverse power rule adds 1 to the exponent, giving 5/2, then divides by the new exponent, and dividing by 5/2 is the same as multiplying by 2/5. Written with radicals, the answer is (2/5)x^2 times the square root of x.

\int x^{3/2}\,dx = \frac{2}{5}x^{5/2} + C

The power rule does not care that the exponent is a fraction

One rule covers every exponent except $-1$ , and $\frac{3}{2}$ is nowhere near that exception. Add 1 to the exponent, then divide by what you get.

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

Here $n + 1 = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}$ , and dividing by $\frac{5}{2}$ is multiplying by its reciprocal.

\int x^{3/2}\,dx = \frac{x^{5/2}}{\frac{5}{2}} + C = \frac{2}{5}x^{5/2} + C

Checking, and reading the answer in root form

Differentiating puts the $\frac{5}{2}$ back out front, where it meets the $\frac{2}{5}$ and cancels.

\frac{d}{dx}\left[\frac{2}{5}x^{5/2}\right] = \frac{2}{5}\cdot\frac{5}{2}x^{5/2-1} = x^{3/2}

Some problems state the integrand as $x\sqrt{x}$ instead. That is the same function, since $x\cdot x^{1/2} = x^{3/2}$ , and the answer can be written $\frac{2}{5}x^2\sqrt{x}$ . Fractional exponents are faster to differentiate, so convert radicals before integrating and convert back only if the problem asks.

Watch the domain

$x^{3/2}$ needs $x \ge 0$ , so this antiderivative describes area only on $[0, \infty)$ . A definite integral with a negative lower limit is not defined for this integrand.

The mistake students make

The most common error is adding 1 to the numerator of the exponent alone, turning $\frac{3}{2}$ into $\frac{4}{2} = 2$ and producing $\frac{x^2}{2}$ . Adding 1 to a fraction means adding $\frac{2}{2}$ , so $\frac{3}{2}$ becomes $\frac{5}{2}$ .

The second is dividing by the old exponent rather than the new one, which gives $\frac{2}{3}x^{5/2}$ . The power rule always divides by the exponent that appears in the answer, not the one you started with.

One derivative catches both

Differentiate whatever you wrote. If the exponent does not drop back to $\frac{3}{2}$ and the coefficient does not cancel to 1, the arithmetic on the exponent went wrong.

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

Where does the 2/5 come from?

It is the reciprocal of the new exponent. The rule divides by $\frac{5}{2}$ , and dividing by $\frac{5}{2}$ is multiplying by $\frac{2}{5}$ .

How do I integrate x times the square root of x?

Rewrite it first: $x\sqrt{x} = x\cdot x^{1/2} = x^{3/2}$ , so the answer is $\frac{2}{5}x^{5/2} + C$ . Radicals almost always become fractional exponents before any integration rule applies.

Does the power rule ever fail for fractional exponents?

No. The only exponent it cannot handle is $n = -1$ , where $n+1 = 0$ and the division is impossible. That single case gives $\int x^{-1}\,dx = \ln|x| + C$ .