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

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

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

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

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

Solution steps

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

Factor

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

= 3 \cdot \frac{1}{6} \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}{6} (- \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

\int 2 x e^{3 x} d x

\frac{d}{d x} (cos^{2} (2 x - 1))

f (t) = t^{5}

x \to 1 lim (\frac{x ^{2} - 1}{x})

x + y^{2} = 56, x + y = 0

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