integral from -infinity to 0 of 1/(6-2x)

Let's compute the integral:

\int_{-∞}^{0} \frac{1}{6 - 2 x} d x

Let

u = 6 - 2 x

. Then

du = - 2 dx

, so

dx = - \frac{du}{2}

\int \frac{1}{6 - 2 x} d x = - \frac{1}{2} \cdot \int \frac{1}{u} d u

We use

\int \frac{1}{u} d u = \ln | u |

. The absolute value covers both signs of the denominator, on any interval where

6 - 2 x \neq 0

Substitute

u

6 - 2 x

back into the antiderivative

ⓘ The derivative of

- \frac{1}{2} \cdot \ln | 6 - 2 x |

\frac{1}{6 - 2 x}

, so

- \frac{1}{2} \cdot \ln | 6 - 2 x |

is an antiderivative of

\frac{1}{6 - 2 x}

\int \frac{1}{6 - 2 x} d x

- \frac{1}{2} \cdot \ln | 6 - 2 x |

- \frac{1}{2} \cdot \ln | 6 - 2 x |

ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{1}{2} \cdot \ln | 6 - 2 x |

- \frac{1 \cdot \ln | 6 - 2 x |}{2}

- \frac{\ln | 6 - 2 x |}{2}

▷

- \frac{\ln | 6 - 2 x |}{2}

The bound

-∞

being infinite, the antiderivative takes its limit there

Let's evaluate the limit of

- \frac{\ln | 6 - 2 x |}{2}

when

x \to -∞

ⓘ A quantity that grows without bound multiplied by a quantity not tending to 0 still grows without bound, the signs giving +∞ or -∞

The limit of

- 2 x

+∞

when

x \to -∞

ⓘ A quantity that grows without bound plus a finite or bounded quantity still grows without bound

The limit of

6 - 2 x

+∞

when

x \to -∞

ⓘ This is a reference limit of the function abs

The limit of

| 6 - 2 x |

+∞

when

x \to -∞

ⓘ This is a reference limit of the function ln

The limit of

\ln | 6 - 2 x |

+∞

when

x \to -∞

ⓘ A quantity that grows without bound divided by a nonzero number still grows without bound, the signs giving +∞ or -∞

So the limit of

- \frac{\ln | 6 - 2 x |}{2}

-∞

when

x \to -∞

The integral is the antiderivative at the upper bound minus at the lower bound:

- \frac{\ln (6)}{2}

- (

-∞

) =

+∞

, so it diverges to

+∞

▶Integral:

+∞