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

Solution

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2 - u + 2} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int - 2 - v^{2} + 2 d v

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 2 cos^{2} (w) d w

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{2} (- 2 \cdot 2 \cdot \int \frac{1 + cos ( 2 w )}{2} d w)

Take the constant out:

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

= \frac{1}{2} (- 2 \cdot 2 \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w)

Apply the Sum Rule:

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

= \frac{1}{2} (- 2 \cdot 2 \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w))

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= \frac{1}{2} (- 2 \cdot 2 \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w)))

Substitute back

= \frac{1}{2} (- 2 \cdot 2 \cdot \frac{1}{2} (arcsin (\frac{1}{2} - (1 + 2 x) + 2) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} - (1 + 2 x) + 2))))

Simplify

\frac{1}{2} (- 2 \cdot 2 \cdot \frac{1}{2} (arcsin (\frac{1}{2} - (1 + 2 x) + 2) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} - (1 + 2 x) + 2)))) : - arcsin (\frac{1}{2} (- 2 x + 1)) - \frac{1}{2} sin (2 arcsin (\frac{- 2 x + 1}{2}))

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

Add a constant to the solution

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

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