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

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

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

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

8 (\frac{1}{4} arctan (\frac{- x + 1}{x + 1}) + \frac{1}{32} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{16} - x + 1 x + 1 + \frac{1}{16} \frac{- x + 1}{x + 1} (x + 1)^{2}) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int \frac{4 v ^{2} ( v ^{2} - 1 )}{( v ^{2} + 1 ) ^{\frac{5}{2}} v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot 4 \cdot \int \frac{v ^{2} ( v ^{2} - 1 )}{( v ^{2} + 1 ) ^{\frac{5}{2}} v ^{2} + 1} d v

(v^{2} + 1)^{\frac{5}{2}} = (v^{2} + 1)^{2} v^{2} + 1

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

(v^{2} + 1)^{2} v^{2} + 1 v^{2} + 1 = (v^{2} + 1)^{3}

= 2 \cdot 4 \cdot \int \frac{v ^{2} ( v ^{2} - 1 )}{( v ^{2} + 1 ) ^{3}} d v

Expand

\frac{v ^{2} ( v ^{2} - 1 )}{( v ^{2} + 1 ) ^{3}} : \frac{v ^{4}}{( v ^{2} + 1 ) ^{3}} - \frac{v ^{2}}{( v ^{2} + 1 ) ^{3}}

= 2 \cdot 4 \cdot \int \frac{v ^{4}}{( v ^{2} + 1 ) ^{3}} - \frac{v ^{2}}{( v ^{2} + 1 ) ^{3}} 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 4 (\int \frac{v ^{4}}{( v ^{2} + 1 ) ^{3}} d v - \int \frac{v ^{2}}{( v ^{2} + 1 ) ^{3}} d v)

\int \frac{v ^{4}}{( v ^{2} + 1 ) ^{3}} d v = \frac{3}{8} arctan (v) - \frac{5}{8} \cdot \frac{v}{1 + v ^{2}} + \frac{v}{4 ( v ^{2} + 1 ) ^{2}}

\int \frac{v ^{2}}{( v ^{2} + 1 ) ^{3}} d v = \frac{1}{8} (arctan (v) - \frac{1}{4} sin (4 arctan (v)))

= 2 \cdot 4 (\frac{3}{8} arctan (v) - \frac{5}{8} \cdot \frac{v}{1 + v ^{2}} + \frac{v}{4 ( v ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (v) - \frac{1}{4} sin (4 arctan (v))))

Substitute back

= 2 \cdot 4 \frac{3}{8} arctan (\frac{1 - x}{1 + x}) - \frac{5}{8} \cdot \frac{\frac{1 - x}{1 + x}}{1 + ( \frac{1 - x}{1 + x} ) ^{2}} + \frac{\frac{1 - x}{1 + x}}{4 ( ( \frac{1 - x}{1 + x} ) ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} sin (4 arctan (\frac{1 - x}{1 + x})))

Simplify

2 \cdot 4 \frac{3}{8} arctan (\frac{1 - x}{1 + x}) - \frac{5}{8} \cdot \frac{\frac{1 - x}{1 + x}}{1 + ( \frac{1 - x}{1 + x} ) ^{2}} + \frac{\frac{1 - x}{1 + x}}{4 ( ( \frac{1 - x}{1 + x} ) ^{2} + 1 ) ^{2}} - \frac{1}{8} (arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} sin (4 arctan (\frac{1 - x}{1 + x}))) : 8 (\frac{1}{4} arctan (\frac{- x + 1}{x + 1}) + \frac{1}{32} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{16} - x + 1 x + 1 + \frac{1}{16} \frac{- x + 1}{x + 1} (x + 1)^{2})

= 8 (\frac{1}{4} arctan (\frac{- x + 1}{x + 1}) + \frac{1}{32} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{16} - x + 1 x + 1 + \frac{1}{16} \frac{- x + 1}{x + 1} (x + 1)^{2})

Add a constant to the solution

= 8 (\frac{1}{4} arctan (\frac{- x + 1}{x + 1}) + \frac{1}{32} sin (4 arctan (\frac{- x + 1}{x + 1})) - \frac{5}{16} - x + 1 x + 1 + \frac{1}{16} \frac{- x + 1}{x + 1} (x + 1)^{2}) + C

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