Integral of $\ln\left(x\right)$

Q: What is the integral of ln(x)?

The integral of ln ≤ft(x) is x ln ≤ft(x) - x + C.

The integral of $\ln\left(x\right)$ is $x \ln\left(x\right) - x + C$ .

\int \ln\left(x\right)\,dx = x \ln\left(x\right) - x + C

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Graph of f(x) = ln(x) with the area from 1 to 2 shaded. The shaded area equals the definite integral, about 0.38629436.

How to work it out

1
We want an antiderivative of f(x) = ln(x).
$\int \ln\left(x\right)\,dx$
2
Method: integration by parts with u = ln x, dv = dx.
3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
$\int \ln\left(x\right)\,dx = x \ln\left(x\right) - x + C$
4
Check by differentiating the answer; you must get the original function back:
$\frac{d}{dx}\left[x \ln\left(x\right) - x\right] = \ln\left(x\right) + x \frac{1}{x} - 1$
5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 1 and 2:
$\int_{1}^{2} \ln\left(x\right)\,dx = \Big[x \ln\left(x\right) - x\Big]_{1}^{2} \approx 0.38629436$

LaTeX source of the formulas on this page

\[ \int \ln\left(x\right)\,dx = x \ln\left(x\right) - x + C \]
\[ \int \ln\left(x\right)\,dx \]
\[ \frac{d}{dx}\left[x \ln\left(x\right) - x\right] = \ln\left(x\right) + x \frac{1}{x} - 1 \]
\[ \int_{1}^{2} \ln\left(x\right)\,dx = \Big[x \ln\left(x\right) - x\Big]_{1}^{2} \approx 0.38629436 \]

Definite integrals of the same function

Interval	Value
from 1 to 2	0.38629436
from 1 to 1.5	0.10819766

Questions

What is the integral of ln(x)?

The integral of $\ln\left(x\right)$ is $x \ln\left(x\right) - x + C$ .

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 ln(x) from 1 to 2?

About 0.38629436.

Integral of ln⁡(x)\ln\left(x\right)ln(x)

How to work it out

Definite integrals of the same function

Questions

More integrals

Integral of $\ln\left(x\right)$