What is the integral of (3-x)^{-1/2} ?

The integral of (3-x)^{-1/2} is x/((3-x)^{1/2)}+(x-6)/((3-x)^{1/2)}+C

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integral of (3-x)^{-1/2}

\int (3 - x)^{- \frac{1}{2}} d x

\frac{x}{( 3 - x ) ^{\frac{1}{2}}} + \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}} + C

Solution steps

\int (3 - x)^{- \frac{1}{2}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 3 - x ) ^{\frac{1}{2}}} d x

Apply Integration By Parts

: \frac{x}{( 3 - x ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x

= \frac{x}{( 3 - x ) ^{\frac{1}{2}}} - \int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x

\int \frac{x}{2 ( 3 - x ) ^{\frac{3}{2}}} d x = - \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}}

= \frac{x}{( 3 - x ) ^{\frac{1}{2}}} - (- \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}})

Simplify

= \frac{x}{( 3 - x ) ^{\frac{1}{2}}} + \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{( 3 - x ) ^{\frac{1}{2}}} + \frac{x - 6}{( 3 - x ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (x^{2} + x + 4)

\frac{d h}{d t} = 4 - 16 t^{2}, h (1) = 0

\frac{d}{d x} (ln (x^{2} - 16 x))

\int \frac{x ^{2} + 3 x - 4}{x ^{2} - 2 x - 8} d x

\int (y^{2} - 1) d y

What is the integral of (3-x)^{-1/2} ?
The integral of (3-x)^{-1/2} is x/((3-x)^{1/2)}+(x-6)/((3-x)^{1/2)}+C