What is the integral of 1/(25-4x^2) ?

The integral of 1/(25-4x^2) is 1/20 (ln| 2/5 x+1|-ln| 2/5 x-1|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(25-4x^2)

\int \frac{1}{25 - 4 x ^{2}} d x

\frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

Solution steps

\int \frac{1}{25 - 4 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{10 ( - 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}{10} \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}{10} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{2}{5} x

= \frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2})

Simplify

\frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2}) : \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

Add a constant to the solution

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

\frac{\partial}{\partial y} (4 x (4 + y)^{- 1})

\frac{6 x}{x + 5}

x \to 8 lim (\frac{( x - 8 )}{x - 8})

\int_{0}^{0.5} 3000 x d x

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

What is the integral of 1/(25-4x^2) ?
The integral of 1/(25-4x^2) is 1/20 (ln| 2/5 x+1|-ln| 2/5 x-1|)+C