What is the integral of 14-(20-3sqrt(16-2x)) ?

The integral of 14-(20-3sqrt(16-2x)) is -6x-(16-2x)^{3/2}+C

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integral of 14-(20-3sqrt(16-2x))

\int 14 - (20 - 3 16 - 2 x) d x

- 6 x - (16 - 2 x)^{\frac{3}{2}} + C

Solution steps

\int 14 - (20 - 3 16 - 2 x) d x

Apply the Sum Rule:

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

= \int 14 d x - \int 20 - 3 16 - 2 x d x

\int 14 d x = 14 x

\int 20 - 3 16 - 2 x d x = 20 x + (16 - 2 x)^{\frac{3}{2}}

= 14 x - (20 x + (16 - 2 x)^{\frac{3}{2}})

Simplify

= - 6 x - (16 - 2 x)^{\frac{3}{2}}

Add a constant to the solution

= - 6 x - (16 - 2 x)^{\frac{3}{2}} + C

y^{- 1}

\frac{\partial}{\partial y} (\frac{1}{y ^{2}})

\frac{d y}{d x} = 12 - y

x^{- 3}

\int - π + 2 t cos (n) t d t

What is the integral of 14-(20-3sqrt(16-2x)) ?
The integral of 14-(20-3sqrt(16-2x)) is -6x-(16-2x)^{3/2}+C