What is the integral of 1/(1-5t) ?

The integral of 1/(1-5t) is -1/5 ln|1-5t|+C

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integral of 1/(1-5t)

\int \frac{1}{1 - 5 t} d t

- \frac{1}{5} ln ∣ 1 - 5 t ∣ + C

Solution steps

\int \frac{1}{1 - 5 t} d t

Apply u-substitution

: \int - \frac{1}{5 u} d u

= \int - \frac{1}{5 u} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{5} ln ∣ u ∣

Substitute back

u = 1 - 5 t

= - \frac{1}{5} ln ∣ 1 - 5 t ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ 1 - 5 t ∣ + C

\frac{\partial}{\partial x} (x^{3} \cdot 4 x + 5 y)

y = x (x - 4)

\frac{d}{d x} (sin (x) + cos (2 x))

\int_{0}^{2} ∣ x - 1 ∣ d x

\int_{0}^{\infty} \frac{5}{x ^{2} + 4} d x