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

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

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integral of sqrt((x-1)/(x+1)) 1/(x^2)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

= \int \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 4 \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{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

= 4 (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

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

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

= 4 (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x - 1}{x + 1}

= 4 arctan (\frac{x - 1}{x + 1}) - \frac{1}{2} arctan (\frac{x - 1}{x + 1}) + \frac{\frac{x - 1}{x + 1}}{1 + ( \frac{x - 1}{x + 1} ) ^{2}}

Simplify

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

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

Add a constant to the solution

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

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