What is the integral of 10-(14-2(16-2x)^{1/2}) ?

The integral of 10-(14-2(16-2x)^{1/2}) is -2/3 (16-2x)^{3/2}+2(16-2x)+C

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integral of 10-(14-2(16-2x)^{1/2})

\int 10 - (14 - 2 (16 - 2 x)^{\frac{1}{2}}) d x

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

Solution steps

\int 10 - (14 - 2 (16 - 2 x)^{\frac{1}{2}}) d x

Apply u-substitution

: \int - u^{\frac{1}{2}} + 2 d u

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

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

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

\int 2 d u = 2 u

= - \frac{2}{3} u^{\frac{3}{2}} + 2 u

Substitute back

u = 16 - 2 x

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

Add a constant to the solution

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

\int 8 x^{9} e^{- x^{5}} d x

\int x^{4} - 8 x^{2} + 16 d x

f (x) = 3 x + 2 x

x \to - 3 lim (\frac{x}{( x + 3 ) ^{4}})

x^{4} + \frac{2}{x}, at x = 1

What is the integral of 10-(14-2(16-2x)^{1/2}) ?
The integral of 10-(14-2(16-2x)^{1/2}) is -2/3 (16-2x)^{3/2}+2(16-2x)+C