What is the integral of (-2x)/((2x+4)-sqrt(4x+8)) ?

The integral of (-2x)/((2x+4)-sqrt(4x+8)) is -x-2sqrt(x+2)+2ln(2|sqrt(x+2)-1|)-1+C

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integral of (-2x)/((2x+4)-sqrt(4x+8))

\int \frac{- 2 x}{( 2 x + 4 ) - 4 x + 8} d x

- x - 2 x + 2 + 2 ln (2 x + 2 - 1) - 1 + C

Solution steps

\int \frac{- 2 x}{2 x + 4 - 4 x + 8} d x

Take the constant out:

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

= - 2 \cdot \int \frac{x}{2 x + 4 - 4 x + 8} d x

Apply u-substitution

= - 2 \cdot \int \frac{u ^{2} - 8}{4 ( u - 2 )} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{4} \cdot \int \frac{u ^{2} - 8}{u - 2} d u

\frac{u ^{2} - 8}{u - 2} = \frac{( u ^{2} - 8 ) ( u + 2 )}{( u ) ^{2} - ( 2 ) ^{2}}

= - 2 \cdot \frac{1}{4} \cdot \int \frac{( u ^{2} - 8 ) ( u + 2 )}{u ^{2} - 2 ^{2}} d u

Apply u-substitution

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

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

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

\int \frac{v ^{2} - 5 v}{v - 4} d v = \frac{( v - 4 ) ^{2}}{2} + 3 (v - 4) - 4 ln ∣ v - 4 ∣

\int 1 d v = v

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

Substitute back

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

Simplify

- 2 \cdot \frac{1}{4} (\frac{( 4 x + 8 + 2 - 4 ) ^{2}}{2} + 3 (4 x + 8 + 2 - 4) - 4 ln 4 x + 8 + 2 - 4 + 4 x + 8 + 2) : - x - 2 x + 2 + 2 ln (2 x + 2 - 1) - 1

= - x - 2 x + 2 + 2 ln (2 x + 2 - 1) - 1

Add a constant to the solution

= - x - 2 x + 2 + 2 ln (2 x + 2 - 1) - 1 + C

What is the integral of (-2x)/((2x+4)-sqrt(4x+8)) ?
The integral of (-2x)/((2x+4)-sqrt(4x+8)) is -x-2sqrt(x+2)+2ln(2|sqrt(x+2)-1|)-1+C