What is the integral of (3x-5)^{1/2} ?

The integral of (3x-5)^{1/2} is 2/9 (3x-5)^{3/2}+C

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integral of (3x-5)^{1/2}

\int (3 x - 5)^{\frac{1}{2}} d x

\frac{2}{9} (3 x - 5)^{\frac{3}{2}} + C

Solution steps

\int (3 x - 5)^{\frac{1}{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x - 5

= \frac{1}{3} \cdot \frac{2}{3} (3 x - 5)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (3 x - 5)^{\frac{3}{2}} : \frac{2}{9} (3 x - 5)^{\frac{3}{2}}

= \frac{2}{9} (3 x - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (3 x - 5)^{\frac{3}{2}} + C

y^{^{''}} = 6 x - 12

y = 6 x^{4} - 6 x^{2}, y = 3 x^{2}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 2 x = sin (2 t)

\frac{d y}{d x} = \frac{1}{2 cos ^{2} ( x )}

\frac{d y}{d x} = y (1 - 2 y)

What is the integral of (3x-5)^{1/2} ?
The integral of (3x-5)^{1/2} is 2/9 (3x-5)^{3/2}+C