What is the integral of 1/(9-4x^2) ?

The integral of 1/(9-4x^2) is 1/12 (ln| 2/3 x+1|-ln| 2/3 x-1|)+C

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integral of 1/(9-4x^2)

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

\frac{1}{12} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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}

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

Substitute back

u = \frac{2}{3} x

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

Simplify

\frac{1}{6} (\frac{ln \frac{2}{3} x + 1}{2} - \frac{ln \frac{2}{3} x - 1}{2}) : \frac{1}{12} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

= \frac{1}{12} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1)

Add a constant to the solution

= \frac{1}{12} (ln \frac{2}{3} x + 1 - ln \frac{2}{3} x - 1) + C

\int_{0}^{\infty} f (a + x) d x

n = 0 \sum \infty \frac{3}{( 1 + i ) ^{n}}

\frac{d y}{d x}, e^{y} = x

\frac{d}{d u} (u sin (v))

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

What is the integral of 1/(9-4x^2) ?
The integral of 1/(9-4x^2) is 1/12 (ln| 2/3 x+1|-ln| 2/3 x-1|)+C