Integral of x ln x

Q: How can I check the answer?

Differentiate it: the derivative of \frac{1}{2}x^{2} \ln\left(x\right) - \frac{1}{4}x^{2} is x \ln\left(x\right). CalcViz checks every antiderivative this way before showing it.

\int x \ln\left(x\right)\,dx = \frac{1}{2}x^{2} \ln\left(x\right) - \frac{1}{4}x^{2} + C

Integration by parts with u = ln x and dv = x dx, which leaves ∫x/2 dx.

Step by step

Integration by parts
∫u dv = uv − ∫v du with u = $\ln\left(x\right)$ (chosen by LIATE) and dv = $x\,dx$ , so du = $\frac{1}{x}\,dx$ and v = $\frac{1}{2}x^{2}$ .
$\int x \ln\left(x\right)\,dx = \frac{1}{2}x^{2} \ln\left(x\right) - \int \frac{1}{2}x\,dx$
Constant multiple rule
Move the constant outside the integral.
$\int \frac{1}{2}x\,dx = \frac{1}{2} \int x\,dx$
Power rule
∫xⁿ dx = xⁿ⁺¹/(n+1).
$\int x\,dx = \frac{1}{2}x^{2}$

f(x) = x ln x
F(x) + C

Constant C = 0.0

Every choice of C gives a valid antiderivative, since shifting up or down doesn’t change the slope.

Open in the integral calculator

Questions students ask

What is the integral of x ln x?

$\int x \ln\left(x\right)\,dx = \frac{1}{2}x^{2} \ln\left(x\right) - \frac{1}{4}x^{2} + C$ .

How do you integrate x ln x?

Integration by parts with u = ln x and dv = x dx, which leaves ∫x/2 dx.

How can I check the answer?

Differentiate it: the derivative of $\frac{1}{2}x^{2} \ln\left(x\right) - \frac{1}{4}x^{2}$ is $x \ln\left(x\right)$ . CalcViz checks every antiderivative this way before showing it.