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

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

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integral of 2+sqrt(4-x)

\int 2 + 4 - x d x

- 2 (4 - x) - \frac{2}{3} (4 - x)^{\frac{3}{2}} + C

Solution steps

\int 2 + 4 - x d x

Apply u-substitution

: \int - 2 - u d u

= \int - 2 - u 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 2 d u - \int u d u

\int 2 d u = 2 u

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

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

Substitute back

u = 4 - x

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

Add a constant to the solution

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

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