What is the integral from-infinity to 0 of 1/(6-4x) ?

The integral from-infinity to 0 of 1/(6-4x) is diverges

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integral from-infinity to 0 of 1/(6-4x)

\int_{- \infty}^{0} \frac{1}{6 - 4 x} d x

diverges

Solution steps

\int_{- \infty}^{0} \frac{1}{6 - 4 x} d x

Compute the indefinite integral:

\int \frac{1}{6 - 4 x} d x = - \frac{1}{4} ln ∣ 6 - 4 x ∣ + C

Compute the boundaries:

\int_{- \infty}^{0} \frac{1}{6 - 4 x} d x =

diverges

= diverges

y^{^{''}} + 7 y^{^{'}} + 14 y = 1096 e^{2 t} cos (8 t)

x \to 4 - lim (\frac{x}{x ^{2} - 16})

\frac{d y}{d x} = \frac{x}{25 y}, y (5) = 1

\frac{d}{d x} (9 x + 1)

cos (t) \cdot e^{t}

What is the integral from-infinity to 0 of 1/(6-4x) ?
The integral from-infinity to 0 of 1/(6-4x) is diverges