What is the integral of 1/(5-4x^2) ?

The integral of 1/(5-4x^2) is 1/(4sqrt(5))(ln| 2/(sqrt(5))x+1|-ln| 2/(sqrt(5))x-1|)+C

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integral of 1/(5-4x^2)

\int \frac{1}{5 - 4 x ^{2}} d x

\frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

Solution steps

\int \frac{1}{5 - 4 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{2 5 ( 1 - u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2 5} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{2 5} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{2}{5} x

= \frac{1}{2 5} \frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2}

Simplify

\frac{1}{2 5} \frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2} : \frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

= \frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

Add a constant to the solution

= \frac{1}{4 5} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

sin (x) + cos (x)

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

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

x \to 0 lim ((1 + 2 x)^{\frac{6}{x}})

n = 2 \sum \infty ln (ln (n))

What is the integral of 1/(5-4x^2) ?
The integral of 1/(5-4x^2) is 1/(4sqrt(5))(ln| 2/(sqrt(5))x+1|-ln| 2/(sqrt(5))x-1|)+C