Integral of
The integral of is .
How to work it out
- 1
We want an antiderivative of f(x) = 1 / (1x + 2).
- 2
Method: ∫1/(kx+b) dx = ln|kx+b|/k.
- 3
That gives the antiderivative; add the constant C because any constant disappears when you differentiate:
- 4
Check by differentiating the answer; you must get the original function back:
- 5
As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:
LaTeX source of the formulas on this page
\[ \int \frac{1}{1 x + 2}\,dx = \frac{\ln\left(\left|1 x + 2\right|\right)}{1} + C \]
\[ \int \frac{1}{1 x + 2}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|1 x + 2\right|\right)}{1}\right] = \frac{\operatorname{sign}\left(x + 2\right)}{\left|x + 2\right|} \]
\[ \int_{0}^{1} \frac{1}{1 x + 2}\,dx = \Big[\frac{\ln\left(\left|1 x + 2\right|\right)}{1}\Big]_{0}^{1} \approx 0.40546511 \]Definite integrals of the same function
| Interval | Value |
|---|---|
| from 0 to 1 | 0.40546511 |
| from 0 to 0.5 | 0.22314355 |
Questions
What is the integral of 1 / (1x + 2)?
The integral of is .
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 1 / (1x + 2) from 0 to 1?
About 0.40546511.