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

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

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integral of sqrt((x-1)/(x+1))

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

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

Solution steps

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

Simplify

\frac{x - 1}{x + 1} : \frac{x - 1}{x + 1}

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

Apply u-substitution

: \int 2 u^{2} - 2 d u

= \int 2 u^{2} - 2 d u

Take the constant out:

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

= 2 \cdot \int u^{2} - 2 d u

Apply Trigonometric Substitution

: \int 2 tan^{2} (v) sec (v) d v

= 2 \cdot \int 2 tan^{2} (v) sec (v) d v

Take the constant out:

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

= 2 \cdot 2 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 2 \cdot 2 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 2 \cdot 2 \cdot \int - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

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

= 2 \cdot 2 (- \int sec (v) d v + \int sec^{3} (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

\int sec^{3} (v) d v = \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

= 2 \cdot 2 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 2 \cdot 2 (- ln tan (arcsec (\frac{1}{2} x + 1)) + sec (arcsec (\frac{1}{2} x + 1)) + \frac{1}{2} sec (arcsec (\frac{1}{2} x + 1)) tan (arcsec (\frac{1}{2} x + 1)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x + 1)) + sec (arcsec (\frac{1}{2} x + 1)))

Simplify

2 \cdot 2 (- ln tan (arcsec (\frac{1}{2} x + 1)) + sec (arcsec (\frac{1}{2} x + 1)) + \frac{1}{2} sec (arcsec (\frac{1}{2} x + 1)) tan (arcsec (\frac{1}{2} x + 1)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x + 1)) + sec (arcsec (\frac{1}{2} x + 1))) : - 2 ln \frac{1}{2} (x - 1) + \frac{1}{2} x + 1 + x - 1 x + 1

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

Add a constant to the solution

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

e^{- x}

5 cos (π x)

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

\frac{d}{d t} (e^{- a t} cos (x t))

y^{^{'}} = \frac{2 x ^{5} e ^{\frac{y}{x}} + x ^{3} y ^{2}}{x ^{4} y}

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