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

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

Graph of f(x) = 1 / (3x + 3) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.60.10.20.30.40.50.60.70.8f(x)
Graph of f(x) = 1 / (3x + 3) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.23104906.

How to work it out

  1. 1

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

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

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

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

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

    ∫0113x+3 dx=[ln⁡(∣3x+3∣)3]01≈0.23104906\int_{0}^{1} \frac{1}{3 x + 3}\,dx = \Big[\frac{\ln\left(\left|3 x + 3\right|\right)}{3}\Big]_{0}^{1} \approx 0.23104906
LaTeX source of the formulas on this page
\[ \int \frac{1}{3 x + 3}\,dx = \frac{\ln\left(\left|3 x + 3\right|\right)}{3} + C \]
\[ \int \frac{1}{3 x + 3}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|3 x + 3\right|\right)}{3}\right] = \frac{3 \frac{3 \operatorname{sign}\left(3 x + 3\right)}{\left|3 x + 3\right|}}{9} \]
\[ \int_{0}^{1} \frac{1}{3 x + 3}\,dx = \Big[\frac{\ln\left(\left|3 x + 3\right|\right)}{3}\Big]_{0}^{1} \approx 0.23104906 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.23104906
from 0 to 0.50.13515504

Questions

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

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

About 0.23104906.

More integrals

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