What is the integral of-1/t ?

The integral of-1/t is -ln|t|+C

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integral of-1/t

\int - \frac{1}{t} d t

- ln ∣ t ∣ + C

Solution steps

\int - \frac{1}{t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= - ln ∣ t ∣

Add a constant to the solution

= - ln ∣ t ∣ + C

\frac{d}{d x} ((sin (3 x))^{2})

f (x) = x x - x^{2}

e^{x}, ln (12)

\frac{d}{d x} (x^{2} sin (\frac{1}{x ^{2}}))

x \to 0 lim (\frac{9}{x})