What is the integral of 1/(y^2-9) ?

The integral of 1/(y^2-9) is -1/6 (ln| y/3+1|-ln| y/3-1|)+C

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integral of 1/(y^2-9)

\int \frac{1}{y ^{2} - 9} d y

- \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1) + C

Solution steps

\int \frac{1}{y ^{2} - 9} d y

Apply Integral Substitution

= \int \frac{1}{3 ( u ^{2} - 1 )} 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 ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{3} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{y}{3}

= \frac{1}{3} (- (\frac{ln \frac{y}{3} + 1}{2} - \frac{ln \frac{y}{3} - 1}{2}))

Simplify

\frac{1}{3} (- (\frac{ln \frac{y}{3} + 1}{2} - \frac{ln \frac{y}{3} - 1}{2})) : - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1)

= - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{y}{3} + 1 - ln \frac{y}{3} - 1) + C

d er i v a t i v e y (θ) = π^{4} cos (θ)

\frac{d}{d t} (9 + t)

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

d er i v a t i v e x^{2} - 2 x + 8

\int x^{5} 1 + x^{3} d x

What is the integral of 1/(y^2-9) ?
The integral of 1/(y^2-9) is -1/6 (ln| y/3+1|-ln| y/3-1|)+C