What is the integral of 3/(9y^2-16) ?

The integral of 3/(9y^2-16) is -1/8 (ln| 3/4 y+1|-ln| 3/4 y-1|)+C

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integral of 3/(9y^2-16)

\int \frac{3}{9 y ^{2} - 16} d y

- \frac{1}{8} (ln \frac{3}{4} y + 1 - ln \frac{3}{4} y - 1) + C

Solution steps

\int \frac{3}{9 y ^{2} - 16} 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}{9 y ^{2} - 16} d y

Apply Integral Substitution

= 3 \cdot \int \frac{1}{12 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 3 \cdot \frac{1}{12} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= 3 \cdot \frac{1}{12} \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

= 3 \cdot \frac{1}{12} (- \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}

= 3 \cdot \frac{1}{12} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{4} y

= 3 \cdot \frac{1}{12} (- (\frac{ln \frac{3}{4} y + 1}{2} - \frac{ln \frac{3}{4} y - 1}{2}))

Simplify

3 \cdot \frac{1}{12} (- (\frac{ln \frac{3}{4} y + 1}{2} - \frac{ln \frac{3}{4} y - 1}{2})) : - \frac{1}{8} (ln \frac{3}{4} y + 1 - ln \frac{3}{4} y - 1)

= - \frac{1}{8} (ln \frac{3}{4} y + 1 - ln \frac{3}{4} y - 1)

Add a constant to the solution

= - \frac{1}{8} (ln \frac{3}{4} y + 1 - ln \frac{3}{4} y - 1) + C

t \cdot sin (3 t)

\int \frac{1}{s} d s

y^{^{'}} = \frac{x}{y}, y (0) = - 1

\int 9 x ln (3 x) d x

\int \frac{1}{( t - 4 ) ^{2}} d t

What is the integral of 3/(9y^2-16) ?
The integral of 3/(9y^2-16) is -1/8 (ln| 3/4 y+1|-ln| 3/4 y-1|)+C