What is the integral of (2x)/(sqrt((4-x)^3)) ?

The integral of (2x)/(sqrt((4-x)^3)) is -2(-2sqrt(4-x)-8/(sqrt(4-x)))+C

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integral of (2x)/(sqrt((4-x)^3))

\int \frac{2 x}{( 4 - x ) ^{3}} d x

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

Solution steps

\int \frac{2 x}{( 4 - x ) ^{3}} 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}{( 4 - x ) ^{3}} d x

(4 - x)^{3} = (4 - x)^{\frac{3}{2}},

assuming

(4 - x) \geq 0

= 2 \cdot \int \frac{x}{( 4 - x ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int - \frac{- u + 4}{u ^{\frac{3}{2}}} d u

= 2 \cdot \int - \frac{- u + 4}{u ^{\frac{3}{2}}} d u

Take the constant out:

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

= 2 (- \int \frac{- u + 4}{u ^{\frac{3}{2}}} d u)

Expand

\frac{- u + 4}{u ^{\frac{3}{2}}} : - \frac{1}{u ^{\frac{1}{2}}} + \frac{4}{u ^{\frac{3}{2}}}

= 2 (- \int - \frac{1}{u ^{\frac{1}{2}}} + \frac{4}{u ^{\frac{3}{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

= 2 (- (- \int \frac{1}{u ^{\frac{1}{2}}} d u + \int \frac{4}{u ^{\frac{3}{2}}} d u))

\int \frac{1}{u ^{\frac{1}{2}}} d u = 2 u^{\frac{1}{2}}

\int \frac{4}{u ^{\frac{3}{2}}} d u = - \frac{8}{u}

= 2 (- (- 2 u^{\frac{1}{2}} - \frac{8}{u}))

Substitute back

u = 4 - x

= 2 (- (- 2 (4 - x)^{\frac{1}{2}} - \frac{8}{4 - x}))

Simplify

2 (- (- 2 (4 - x)^{\frac{1}{2}} - \frac{8}{4 - x})) : - 2 (- 2 4 - x - \frac{8}{4 - x})

= - 2 (- 2 4 - x - \frac{8}{4 - x})

Add a constant to the solution

= - 2 (- 2 4 - x - \frac{8}{4 - x}) + C

\frac{d}{d x} (5 ln (x^{5}))

(e^{x})^{^{'}}

\int 4 x^{- \frac{6}{7}} d x

x y^{^{'}} + (2 x - 3) y = 4 x^{4}

\frac{( s + 1 )}{( s + 2 ) ( s - 3 )}

What is the integral of (2x)/(sqrt((4-x)^3)) ?
The integral of (2x)/(sqrt((4-x)^3)) is -2(-2sqrt(4-x)-8/(sqrt(4-x)))+C