What is the integral of 1/(1-3y) ?

The integral of 1/(1-3y) is -1/3 ln|1-3y|+C

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integral of 1/(1-3y)

\int \frac{1}{1 - 3 y} d y

- \frac{1}{3} ln ∣ 1 - 3 y ∣ + C

Solution steps

\int \frac{1}{1 - 3 y} d y

Apply u-substitution

: \int - \frac{1}{3 u} d u

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 - 3 y

= - \frac{1}{3} ln ∣ 1 - 3 y ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 1 - 3 y ∣ + C

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)

f (x) = 5 x^{2} - 6

f (x) = 4 x^{2} + 2 x, (- 1, 2)

- \frac{1}{2 x ^{2}} + \frac{32}{5 x ^{5}}

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