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

The integral of (2+sqrt(x))/(2-sqrt(x)) is -x-8sqrt(x)-16ln|-sqrt(x)+2|+20+C

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

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

- x - 8 x - 16 ln - x + 2 + 20 + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - \frac{( - v + 2 ) ( - v + 4 )}{v} d v

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

Take the constant out:

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

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

Expand

\frac{( - v + 2 ) ( - v + 4 )}{v} : v - 6 + \frac{8}{v}

= 2 (- \int v - 6 + \frac{8}{v} 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 (- (\int v d v - \int 6 d v + \int \frac{8}{v} d v))

\int v d v = \frac{v ^{2}}{2}

\int 6 d v = 6 v

\int \frac{8}{v} d v = 8 ln ∣ v ∣

= 2 (- (\frac{v ^{2}}{2} - 6 v + 8 ln ∣ v ∣))

Substitute back

= 2 (- (\frac{( 2 - x ) ^{2}}{2} - 6 (2 - x) + 8 ln 2 - x))

Simplify

2 (- (\frac{( 2 - x ) ^{2}}{2} - 6 (2 - x) + 8 ln 2 - x)) : - x - 8 x - 16 ln - x + 2 + 20

= - x - 8 x - 16 ln - x + 2 + 20

Add a constant to the solution

= - x - 8 x - 16 ln - x + 2 + 20 + C

\int_{- 2}^{9} \frac{7}{( 4 x - 3 ) ^{2}} d x

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2 x e^{x}

\int 2 t e^{- t^{2}} d t

What is the integral of (2+sqrt(x))/(2-sqrt(x)) ?
The integral of (2+sqrt(x))/(2-sqrt(x)) is -x-8sqrt(x)-16ln|-sqrt(x)+2|+20+C