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

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

integral

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

y = 4 t + 2

\frac{2}{( x ^{2} - 9 )}

\frac{1}{2 x + 1}

(e^{e})^{x} \cdot e^{2} x

\frac{( 1 - e ^{- x} )}{x}

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