What is the integral from-infinity to 0 of 1/(6-8x) ?

The integral from-infinity to 0 of 1/(6-8x) is diverges

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integral from-infinity to 0 of 1/(6-8x)

\int_{- \infty}^{0} \frac{1}{6 - 8 x} d x

diverges

Solution steps

\int_{- \infty}^{0} \frac{1}{6 - 8 x} d x

Compute the indefinite integral:

\int \frac{1}{6 - 8 x} d x = - \frac{1}{8} ln ∣ 6 - 8 x ∣ + C

Compute the boundaries:

\int_{- \infty}^{0} \frac{1}{6 - 8 x} d x =

diverges

= diverges

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y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{( 13.5 e ^{(3 t)} )}{t ^{2} + 1}

\frac{d}{d x} (cosh (ln (e^{x^{3}})))

\int \frac{9}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{9}{x} d x

What is the integral from-infinity to 0 of 1/(6-8x) ?
The integral from-infinity to 0 of 1/(6-8x) is diverges