What is the integral of 1/(4-3x^2) ?

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

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integral of 1/(4-3x^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{3}{2} x

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

Simplify

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

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

Add a constant to the solution

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

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