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Integral of 12x+4\frac{1}{2 x + 4}

The integral of 12x+4\frac{1}{2 x + 4} is ln⁡(∣2x+4∣)2+C\frac{\ln\left(\left|2 x + 4\right|\right)}{2} + C.

Graph of f(x) = 1 / (2x + 4) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.60.150.20.250.30.35f(x)
Graph of f(x) = 1 / (2x + 4) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.20273255.

How to work it out

  1. 1

    We want an antiderivative of f(x) = 1 / (2x + 4).

    ∫12x+4 dx\int \frac{1}{2 x + 4}\,dx
  2. 2

    Method: ∫1/(kx+b) dx = ln|kx+b|/k.

  3. 3

    That gives the antiderivative; add the constant C because any constant disappears when you differentiate:

    ∫12x+4 dx=ln⁡(∣2x+4∣)2+C\int \frac{1}{2 x + 4}\,dx = \frac{\ln\left(\left|2 x + 4\right|\right)}{2} + C
  4. 4

    Check by differentiating the answer; you must get the original function back:

    ddx[ln⁡(∣2x+4∣)2]=22sign⁡(2x+4)∣2x+4∣4\frac{d}{dx}\left[\frac{\ln\left(\left|2 x + 4\right|\right)}{2}\right] = \frac{2 \frac{2 \operatorname{sign}\left(2 x + 4\right)}{\left|2 x + 4\right|}}{4}
  5. 5

    As a definite integral, evaluate the antiderivative at both ends and subtract. Between 0 and 1:

    ∫0112x+4 dx=[ln⁡(∣2x+4∣)2]01≈0.20273255\int_{0}^{1} \frac{1}{2 x + 4}\,dx = \Big[\frac{\ln\left(\left|2 x + 4\right|\right)}{2}\Big]_{0}^{1} \approx 0.20273255
LaTeX source of the formulas on this page
\[ \int \frac{1}{2 x + 4}\,dx = \frac{\ln\left(\left|2 x + 4\right|\right)}{2} + C \]
\[ \int \frac{1}{2 x + 4}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|2 x + 4\right|\right)}{2}\right] = \frac{2 \frac{2 \operatorname{sign}\left(2 x + 4\right)}{\left|2 x + 4\right|}}{4} \]
\[ \int_{0}^{1} \frac{1}{2 x + 4}\,dx = \Big[\frac{\ln\left(\left|2 x + 4\right|\right)}{2}\Big]_{0}^{1} \approx 0.20273255 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.20273255
from 0 to 0.50.11157178

Questions

What is the integral of 1 / (2x + 4)?

The integral of 12x+4\frac{1}{2 x + 4} is ln⁡(∣2x+4∣)2+C\frac{\ln\left(\left|2 x + 4\right|\right)}{2} + 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 1 / (2x + 4) from 0 to 1?

About 0.20273255.

More integrals

    Integral of 1 / (2x + 4): ln(abs(2x + 4)) / 2 + C (with steps) | WordGPT