What is the integral of 1/(sqrt(x+1)+\sqrt{x-1)} ?

The integral of 1/(sqrt(x+1)+\sqrt{x-1)} is 1/3 (x+1)^{3/2}-1/3 (x-1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x+1)+\sqrt{x-1)}

\int \frac{1}{x + 1 + x - 1} d x

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

Solution steps

\int \frac{1}{x + 1 + x - 1} d x

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{x + 1}{( x + 1 ) ^{2} - ( x - 1 ) ^{2}} d x = \frac{1}{3} (x + 1)^{\frac{3}{2}}

\int \frac{x - 1}{( x + 1 ) ^{2} - ( x - 1 ) ^{2}} d x = \frac{1}{3} (x - 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

f (x) = x^{2} - 1, g (x) = - x + 2, 0, 1

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

\int_{0}^{3} \frac{1}{9 - x ^{2}} d x

\frac{d}{d x} (- A B^{- x} + C)

\frac{d}{d x} (2 x (1 - x)^{3} - 3 x^{2} (1 - x)^{2})

What is the integral of 1/(sqrt(x+1)+\sqrt{x-1)} ?
The integral of 1/(sqrt(x+1)+\sqrt{x-1)} is 1/3 (x+1)^{3/2}-1/3 (x-1)^{3/2}+C