What is the integral of (4+sqrt(x))/(4-sqrt(x)) ?

The integral of (4+sqrt(x))/(4-sqrt(x)) is -x-16sqrt(x)-64ln|-sqrt(x)+4|+C

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integral of (4+sqrt(x))/(4-sqrt(x))

\int \frac{4 + x}{4 - x} d x

- x - 16 x - 64 ln - x + 4 + C

Solution steps

\int \frac{4 + x}{4 - x} d x

Apply u-substitution

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

Long division

\frac{u ( 4 + u )}{4 - u} : - u - 8 + \frac{32}{- u + 4}

= 2 \cdot \int - u - 8 + \frac{32}{- u + 4} 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 u d u - \int 8 d u + \int \frac{32}{- u + 4} d u)

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

\int 8 d u = 8 u

\int \frac{32}{- u + 4} d u = - 32 ln ∣ - u + 4 ∣

= 2 (- \frac{u ^{2}}{2} - 8 u - 32 ln ∣ - u + 4 ∣)

Substitute back

u = x

= 2 (- \frac{( x ) ^{2}}{2} - 8 x - 32 ln - x + 4)

Simplify

2 (- \frac{( x ) ^{2}}{2} - 8 x - 32 ln - x + 4) : - x - 16 x - 64 ln - x + 4

= - x - 16 x - 64 ln - x + 4

Add a constant to the solution

= - x - 16 x - 64 ln - x + 4 + C

What is the integral of (4+sqrt(x))/(4-sqrt(x)) ?
The integral of (4+sqrt(x))/(4-sqrt(x)) is -x-16sqrt(x)-64ln|-sqrt(x)+4|+C