Integral of xln(x)
The integral of xln(x) is 2x2ln(x)−4x2+C.
Graph of f(x) = x * ln(x) with the area from 1 to 2 shaded. The shaded area equals the definite integral, about 0.63629436.How to work it out
- 1
We want an antiderivative of f(x) = x * ln(x).
∫xln(x)dx - 2
Method: integration by parts with u = ln x, dv = x dx.
- 3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
∫xln(x)dx=2x2ln(x)−4x2+C - 4
Check by differentiating the answer; you must get the original function back:
dxd[2x2ln(x)−4x2]=42(2xln(x)+x2x1)−168x - 5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 1 and 2:
∫12xln(x)dx=[2x2ln(x)−4x2]12≈0.63629436
LaTeX source of the formulas on this page
\[ \int x \ln\left(x\right)\,dx = \frac{x^{2} \ln\left(x\right)}{2} - \frac{x^{2}}{4} + C \]
\[ \int x \ln\left(x\right)\,dx \]
\[ \frac{d}{dx}\left[\frac{x^{2} \ln\left(x\right)}{2} - \frac{x^{2}}{4}\right] = \frac{2 \left(2 x \ln\left(x\right) + x^{2} \frac{1}{x}\right)}{4} - \frac{8 x}{16} \]
\[ \int_{1}^{2} x \ln\left(x\right)\,dx = \Big[\frac{x^{2} \ln\left(x\right)}{2} - \frac{x^{2}}{4}\Big]_{1}^{2} \approx 0.63629436 \]Definite integrals of the same function
| Interval | Value |
|---|
| from 1 to 2 | 0.63629436 |
| from 1 to 1.5 | 0.14364825 |
Questions
What is the integral of x * ln(x)?
The integral of xln(x) is 2x2ln(x)−4x2+C.
Why do we add + C?
Every function F(x) + C has the same derivative, so the antiderivative is only defined up to a constant.
What is the integral of x * ln(x) from 1 to 2?
About 0.63629436.