What is the integral of 1/(sqrt(4-(u+3))) ?

The integral of 1/(sqrt(4-(u+3))) is -2sqrt(-u+1)+C

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integral of 1/(sqrt(4-(u+3)))

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

- 2 - u + 1 + C

Solution steps

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

Simplify

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

Apply u-substitution

: \int - \frac{1}{v} d v

= \int - \frac{1}{v} d v

Take the constant out:

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

= - \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= - 2 v^{\frac{1}{2}}

Substitute back

v = - u + 1

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

Apply radical rule:

= - 2 - u + 1

Add a constant to the solution

= - 2 - u + 1 + C

\frac{d}{d x} (2 sin^{2} (\frac{x}{2}))

f (x) = \frac{- 6 x}{x ^{2} + 1}, at x = 0

\int tan (θ) sec (θ) sec^{2} (θ) d θ

\frac{d}{d x} (ln (4 + ln (x)))

\frac{\partial}{\partial x} (a x ln (a x))

What is the integral of 1/(sqrt(4-(u+3))) ?
The integral of 1/(sqrt(4-(u+3))) is -2sqrt(-u+1)+C