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Integral of 14x+1\frac{1}{4 x + 1}

The integral of 14x+1\frac{1}{4 x + 1} is ln⁡(∣4x+1∣)4+C\frac{\ln\left(\left|4 x + 1\right|\right)}{4} + C.

Graph of f(x) = 1 / (4x + 1) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.6-4-224f(x)
Graph of f(x) = 1 / (4x + 1) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.40235948.

How to work it out

  1. 1

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

    ∫14x+1 dx\int \frac{1}{4 x + 1}\,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:

    ∫14x+1 dx=ln⁡(∣4x+1∣)4+C\int \frac{1}{4 x + 1}\,dx = \frac{\ln\left(\left|4 x + 1\right|\right)}{4} + C
  4. 4

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

    ddx[ln⁡(∣4x+1∣)4]=44sign⁡(4x+1)∣4x+1∣16\frac{d}{dx}\left[\frac{\ln\left(\left|4 x + 1\right|\right)}{4}\right] = \frac{4 \frac{4 \operatorname{sign}\left(4 x + 1\right)}{\left|4 x + 1\right|}}{16}
  5. 5

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

    ∫0114x+1 dx=[ln⁡(∣4x+1∣)4]01≈0.40235948\int_{0}^{1} \frac{1}{4 x + 1}\,dx = \Big[\frac{\ln\left(\left|4 x + 1\right|\right)}{4}\Big]_{0}^{1} \approx 0.40235948
LaTeX source of the formulas on this page
\[ \int \frac{1}{4 x + 1}\,dx = \frac{\ln\left(\left|4 x + 1\right|\right)}{4} + C \]
\[ \int \frac{1}{4 x + 1}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|4 x + 1\right|\right)}{4}\right] = \frac{4 \frac{4 \operatorname{sign}\left(4 x + 1\right)}{\left|4 x + 1\right|}}{16} \]
\[ \int_{0}^{1} \frac{1}{4 x + 1}\,dx = \Big[\frac{\ln\left(\left|4 x + 1\right|\right)}{4}\Big]_{0}^{1} \approx 0.40235948 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.40235948
from 0 to 0.50.27465307

Questions

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

The integral of 14x+1\frac{1}{4 x + 1} is ln⁡(∣4x+1∣)4+C\frac{\ln\left(\left|4 x + 1\right|\right)}{4} + 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 / (4x + 1) from 0 to 1?

About 0.40235948.

More integrals

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