What is the integral of wsqrt(w-3) ?

The integral of wsqrt(w-3) is 2/5 (w-3)^{5/2}+2(w-3)^{3/2}+C

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integral of wsqrt(w-3)

\int w w - 3 d w

\frac{2}{5} (w - 3)^{\frac{5}{2}} + 2 (w - 3)^{\frac{3}{2}} + C

Solution steps

\int w w - 3 d w

Apply u-substitution

: \int (u + 3) u d u

= \int (u + 3) u d u

Expand

(u + 3) u : u^{\frac{3}{2}} + 3 u

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}}

Substitute back

u = w - 3

= \frac{2}{5} (w - 3)^{\frac{5}{2}} + 2 (w - 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (w - 3)^{\frac{5}{2}} + 2 (w - 3)^{\frac{3}{2}} + C

\int \frac{( x - 5 )}{x ^{2} + 2 x - 3} d x

\frac{( e ^{x} + e ^{- x} )}{2}

\frac{d}{d x} (\frac{x - 2}{3})

\frac{d}{d x} (- sin^{3} (x))

\int_{0}^{1} 8 x y

What is the integral of wsqrt(w-3) ?
The integral of wsqrt(w-3) is 2/5 (w-3)^{5/2}+2(w-3)^{3/2}+C