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Integral of 15x+2\frac{1}{5 x + 2}

The integral of 15x+2\frac{1}{5 x + 2} is ln⁡(∣5x+2∣)5+C\frac{\ln\left(\left|5 x + 2\right|\right)}{5} + C.

Graph of f(x) = 1 / (5x + 2) with the area from 0 to 1 shaded-0.4-0.20.20.40.60.811.21.41.6-3-2-1123f(x)
Graph of f(x) = 1 / (5x + 2) with the area from 0 to 1 shaded. The shaded area equals the definite integral, about 0.25055259.

How to work it out

  1. 1

    We want an antiderivative of f(x) = 1 / (5x + 2).

    ∫15x+2 dx\int \frac{1}{5 x + 2}\,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:

    ∫15x+2 dx=ln⁡(∣5x+2∣)5+C\int \frac{1}{5 x + 2}\,dx = \frac{\ln\left(\left|5 x + 2\right|\right)}{5} + C
  4. 4

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

    ddx[ln⁡(∣5x+2∣)5]=55sign⁡(5x+2)∣5x+2∣25\frac{d}{dx}\left[\frac{\ln\left(\left|5 x + 2\right|\right)}{5}\right] = \frac{5 \frac{5 \operatorname{sign}\left(5 x + 2\right)}{\left|5 x + 2\right|}}{25}
  5. 5

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

    ∫0115x+2 dx=[ln⁡(∣5x+2∣)5]01≈0.25055259\int_{0}^{1} \frac{1}{5 x + 2}\,dx = \Big[\frac{\ln\left(\left|5 x + 2\right|\right)}{5}\Big]_{0}^{1} \approx 0.25055259
LaTeX source of the formulas on this page
\[ \int \frac{1}{5 x + 2}\,dx = \frac{\ln\left(\left|5 x + 2\right|\right)}{5} + C \]
\[ \int \frac{1}{5 x + 2}\,dx \]
\[ \frac{d}{dx}\left[\frac{\ln\left(\left|5 x + 2\right|\right)}{5}\right] = \frac{5 \frac{5 \operatorname{sign}\left(5 x + 2\right)}{\left|5 x + 2\right|}}{25} \]
\[ \int_{0}^{1} \frac{1}{5 x + 2}\,dx = \Big[\frac{\ln\left(\left|5 x + 2\right|\right)}{5}\Big]_{0}^{1} \approx 0.25055259 \]

Definite integrals of the same function

IntervalValue
from 0 to 10.25055259
from 0 to 0.50.16218604

Questions

What is the integral of 1 / (5x + 2)?

The integral of 15x+2\frac{1}{5 x + 2} is ln⁡(∣5x+2∣)5+C\frac{\ln\left(\left|5 x + 2\right|\right)}{5} + 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 / (5x + 2) from 0 to 1?

About 0.25055259.

More integrals

    Integral of 1 / (5x + 2): ln(abs(5x + 2)) / 5 + C (with steps) | WordGPT