AP Calculus BC

Integral of sqrt(x) ln x: Parts With a Fraction

Q: What does sqrt(x) ln x integrate to?

One application of parts gives \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C.

Q: Which factor should be u?

Take \ln x. Its derivative \frac{1}{x} is simpler than itself, while the other choice forces you to integrate a logarithm before the problem has begun and then to unwind a cyclic integral at the end.

Q: Do I need absolute value bars around ln x?

No. The factor \sqrt{x} already restricts the domain to x > 0, where \ln x is defined, so \ln\left|x\right| would add nothing.

The logarithm is the factor you want to differentiate, so take u equal to ln x and dv equal to the square root of x dx, which leaves a plain power integral behind. The result is two thirds x to the three halves times ln x, minus four ninths x to the three halves, plus C.

\int \sqrt{x}\,\ln x\,dx = \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C

The logarithm is always u

Write the root as a power first: $\sqrt{x} = x^{1/2}$ . Take $u = \ln x$ and $dv = x^{1/2}dx$ , so $du = \frac{dx}{x}$ and $v = \frac{2}{3}x^{3/2}$ . The logarithm goes in $u$ because differentiating it removes it, and there is no reason to integrate it.

\int \sqrt{x}\,\ln x\,dx = \frac{2}{3}x^{3/2}\ln x - \frac{2}{3}\int x^{3/2}\cdot\frac{1}{x}\,dx

The leftover integrand collapses by subtracting exponents: $\frac{x^{3/2}}{x} = x^{1/2}$ . It is an ordinary power, and the reverse power rule finishes it.

\int \sqrt{x}\,\ln x\,dx = \frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C

Fractions everywhere, structure unchanged

Compare this with $\int x\ln x\,dx$ . Nothing about the method changes when the power is $\frac{1}{2}$ instead of $1$ : the same $u$ , the same single application of parts, the same power integral at the end. Only the arithmetic is fussier.

The $\frac{4}{9}$ is two copies of $\frac{2}{3}$ multiplied together, one from $v$ and one from integrating $x^{1/2}$ . Seeing that pattern is a fast way to check the coefficient without redoing the work.

Domain, and what C means here

The integrand only exists for x greater than 0, so the antiderivative lives on a single interval and no absolute value bars are needed around ln x. One constant of integration covers the whole domain, unlike integrals that break at a vertical asymptote.

The mistakes students make

Two of these are arithmetic slips with fractional exponents, and the first is a choice that costs you three extra steps.

Choosing $u = \sqrt{x}$ and $dv = \ln x\,dx$ , which requires knowing $\int \ln x\,dx$ before you can even write $v$ . The leftover integral then contains the original integral again, so you have to recognise the cyclic structure and solve for it algebraically to reach the same answer that $u = \ln x$ hands you directly.
Writing $v = \frac{x^{3/2}}{3}$ by dividing by the numerator of the new exponent instead of the exponent itself. The reverse power rule divides by $\frac{3}{2}$ , which multiplies by $\frac{2}{3}$ .
Stopping at $\frac{2}{3}x^{3/2}\ln x - \frac{2}{3}x^{3/2}$ . That differentiates to $\sqrt{x}\ln x - \frac{\sqrt{x}}{3}$ , so the answer is off by a term.

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 does sqrt(x) ln x integrate to?

One application of parts gives $\frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C$ .

Which factor should be u?

Take $\ln x$ . Its derivative $\frac{1}{x}$ is simpler than itself, while the other choice forces you to integrate a logarithm before the problem has begun and then to unwind a cyclic integral at the end.

Do I need absolute value bars around ln x?

No. The factor $\sqrt{x}$ already restricts the domain to $x > 0$ , where $\ln x$ is defined, so $\ln\left|x\right|$ would add nothing.