What is the integral of ((3(4u+1)^{1/2}))/4 ?

The integral of ((3(4u+1)^{1/2}))/4 is 1/8 (4u+1)^{3/2}+C

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integral of ((3(4u+1)^{1/2}))/4

\int \frac{( 3 ( 4 u + 1 ) ^{\frac{1}{2}} )}{4} d u

\frac{1}{8} (4 u + 1)^{\frac{3}{2}} + C

Solution steps

\int \frac{3 ( 4 u + 1 ) ^{\frac{1}{2}}}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int (4 u + 1)^{\frac{1}{2}} d u

Apply u-substitution

: \int \frac{v ^{\frac{1}{2}}}{4} d v

= \frac{3}{4} \cdot \int \frac{v ^{\frac{1}{2}}}{4} d v

Take the constant out:

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot \int v^{\frac{1}{2}} d v

Apply the Power Rule

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

= \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

v = 4 u + 1

= \frac{3}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 u + 1)^{\frac{3}{2}}

Simplify

\frac{3}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 u + 1)^{\frac{3}{2}} : \frac{1}{8} (4 u + 1)^{\frac{3}{2}}

= \frac{1}{8} (4 u + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{8} (4 u + 1)^{\frac{3}{2}} + C

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

\int \frac{42 w ^{7} - 7 w}{7 w ^{5}} d w

x \to 2 lim (\frac{x ^{3} - 8}{x + 2})

(4, 0), (- 2, 4)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- t}, y (0) = 0, y^{^{'}} (0) = 2

What is the integral of ((3(4u+1)^{1/2}))/4 ?
The integral of ((3(4u+1)^{1/2}))/4 is 1/8 (4u+1)^{3/2}+C