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

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integral of sqrt(1-1/x)

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

2 - \frac{1}{4} ln \frac{x - 1}{x} + 1 - \frac{1}{4 ( \frac{x - 1}{x} + 1 )} + \frac{1}{4} ln \frac{x - 1}{x} - 1 - \frac{1}{4 ( \frac{x - 1}{x} - 1 )} + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{( u ^{2} - 1 ) ^{\frac{3}{2}} u ^{2} - 1} d u

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

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{3}{2}} u ^{2} - 1} d u

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

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

(u^{2} - 1) u^{2} - 1 u^{2} - 1 = (u^{2} - 1)^{2}

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

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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

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

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

= 2 (- \frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4 ( u + 1 )} + \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )})

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

= 2 - \frac{1}{4} ln \frac{x - 1}{x} + 1 - \frac{1}{4 ( \frac{x - 1}{x} + 1 )} + \frac{1}{4} ln \frac{x - 1}{x} - 1 - \frac{1}{4 ( \frac{x - 1}{x} - 1 )} + C

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