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

The integral of 1/((a+b)-(a-b)x^2) is 1/(sqrt(-a+b)\sqrt{a+b)}arctan(sqrt((-a+b)/(a+b))x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((a+b)-(a-b)x^2)

\int \frac{1}{( a + b ) - ( a - b ) x ^{2}} d x

\frac{1}{- a + b a + b} arctan (\frac{- a + b}{a + b} x) + C

Solution steps

\int \frac{1}{a + b - ( a - b ) x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{( u ^{2} + 1 ) - a + b a + b} d u

Take the constant out:

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

= \frac{1}{- a + b a + b} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{- a + b a + b} arctan (u)

Substitute back

u = \frac{- a + b}{a + b} x

= \frac{1}{- a + b a + b} arctan (\frac{- a + b}{a + b} x)

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{- a + b a + b} arctan (\frac{- a + b}{a + b} x)

Add a constant to the solution

= \frac{1}{- a + b a + b} arctan (\frac{- a + b}{a + b} x) + C

What is the integral of 1/((a+b)-(a-b)x^2) ?
The integral of 1/((a+b)-(a-b)x^2) is 1/(sqrt(-a+b)\sqrt{a+b)}arctan(sqrt((-a+b)/(a+b))x)+C