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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u + u + 2} d u

= \int \frac{1}{u + u + 2} d u

Multiply by the conjugate of

u + u + 2 : \frac{- 2}{u - u + 2}

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

Simplify

= (- \frac{1}{2}) \cdot \int \frac{1}{\frac{1}{u - u + 2}} d u

Simplify

= (- \frac{1}{2}) \cdot \int u - u + 2 d u

Apply the Sum Rule:

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

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

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int u + 2 d u = \frac{2}{3} (u + 2)^{\frac{3}{2}}

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = 2 + y - 2 x + 3

\frac{d}{d x} (\frac{2}{( x + 3 ) ^{3}})

y

2

- csc (x) cot (x)

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