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

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

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integral of 3/(1-y)

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

- 3 ln ∣ 1 - y ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- ln ∣ u ∣)

Substitute back

u = 1 - y

= 3 (- ln ∣ 1 - y ∣)

Simplify

= - 3 ln ∣ 1 - y ∣

Add a constant to the solution

= - 3 ln ∣ 1 - y ∣ + C

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

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

n \to \infty lim (\frac{n}{n})

n = 1 \sum \infty \frac{7}{n ^{2} + 36}

x \to 0 lim (∣ x ∣)