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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

= \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{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

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