What is the integral of 3/(9-4x^2) ?

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

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integral of 3/(9-4x^2)

\int \frac{3}{9 - 4 x ^{2}} d x

\frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

Solution steps

\int \frac{3}{9 - 4 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 - 4 x ^{2}} d x

Apply Integral Substitution

= 3 \cdot \int \frac{1}{6 ( - 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}{6} \cdot \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}{6} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{2}{3} x

= 3 \cdot \frac{1}{6} (\frac{ln \frac{2}{3} x + 1}{2} - \frac{ln \frac{2}{3} x - 1}{2})

Simplify

3 \cdot \frac{1}{6} (\frac{ln \frac{2}{3} x + 1}{2} - \frac{ln \frac{2}{3} x - 1}{2}) : \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

= \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

Add a constant to the solution

= \frac{1}{4} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

f (x) = \frac{2 - x ^{2}}{2 + x ^{2}}

f (x)

(- 5, 5), (5, - 5)

(ln (3))^{^{'}}

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