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

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

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

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

ln \frac{1 - x}{1 + x} + 1 + \frac{1}{\frac{1 - x}{1 + x} + 1} - ln \frac{1 - x}{1 + x} - 1 + \frac{1}{\frac{1 - x}{1 + x} - 1} + C

Solution steps

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

= - 4 \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

= - 4 (- \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 )}

= - 4 (- \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 = \frac{1 - x}{1 + x}

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

Simplify

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

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

Add a constant to the solution

= ln \frac{1 - x}{1 + x} + 1 + \frac{1}{\frac{1 - x}{1 + x} + 1} - ln \frac{1 - x}{1 + x} - 1 + \frac{1}{\frac{1 - x}{1 + x} - 1} + C

\frac{d}{d x} (\frac{c x}{1 + c x})

φ^{3} cos (φ^{4})

\int_{0}^{5} (6 x - x^{2})^{2} - x^{2} d x

n = 1 \sum \infty \frac{5}{2 + 3 ^{n}}

\int 2 arcsin (x) d x

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