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Integral of 13x+4\frac{1}{3 x + 4}

The integral of 13x+4\frac{1}{3 x + 4} is ln⁡(∣3x+4∣)3+C\frac{\ln\left(\left|3 x + 4\right|\right)}{3} + C.

Graph of f(x) = 1 / (3x + 4) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.60.10.150.20.250.30.350.40.45f(x)
Graph of f(x) = 1 / (3x + 4) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.1865386.

How to work it out

  1. 1

    We want an antiderivative of f(x) = 1 / (3x + 4).

    ∫13x+4 dx\int \frac{1}{3 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:

    ∫13x+4 dx=ln⁡(∣3x+4∣)3+C\int \frac{1}{3 x + 4}\,dx = \frac{\ln\left(\left|3 x + 4\right|\right)}{3} + C
  4. 4

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

    ddx[ln⁡(∣3x+4∣)3]=33sign⁡(3x+4)∣3x+4∣9\frac{d}{dx}\left[\frac{\ln\left(\left|3 x + 4\right|\right)}{3}\right] = \frac{3 \frac{3 \operatorname{sign}\left(3 x + 4\right)}{\left|3 x + 4\right|}}{9}
  5. 5

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

    ∫0113x+4 dx=[ln⁡(∣3x+4∣)3]01≈0.1865386\int_{0}^{1} \frac{1}{3 x + 4}\,dx = \Big[\frac{\ln\left(\left|3 x + 4\right|\right)}{3}\Big]_{0}^{1} \approx 0.1865386
LaTeX source of the formulas on this page
\[ \int \frac{1}{3 x + 4}\,dx = \frac{\ln\left(\left|3 x + 4\right|\right)}{3} + C \]
\[ \int \frac{1}{3 x + 4}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|3 x + 4\right|\right)}{3}\right] = \frac{3 \frac{3 \operatorname{sign}\left(3 x + 4\right)}{\left|3 x + 4\right|}}{9} \]
\[ \int_{0}^{1} \frac{1}{3 x + 4}\,dx = \Big[\frac{\ln\left(\left|3 x + 4\right|\right)}{3}\Big]_{0}^{1} \approx 0.1865386 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.1865386
from 0 to 0.50.10615124

Questions

What is the integral of 1 / (3x + 4)?

The integral of 13x+4\frac{1}{3 x + 4} is ln⁡(∣3x+4∣)3+C\frac{\ln\left(\left|3 x + 4\right|\right)}{3} + 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 / (3x + 4) from 0 to 1?

About 0.1865386.

More integrals

    Integral of 1 / (3x + 4): ln(abs(3x + 4)) / 3 + C (with steps) | WordGPT