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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{9} (3 x + 1)^{\frac{3}{2}} + C

Solution steps

\int (3 x + 1)^{\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 + 1

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

Simplify

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

= \frac{2}{9} (3 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (3 x + 1)^{\frac{3}{2}} + C

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

f (x) = \frac{2 x}{x ^{2} + 1}, (- 1, - 1)

\frac{2 ( x + 1 )}{x ^{2} + 1}

r^{2} - 9 x^{2}

N - 0.0002 N^{2} = \frac{d N}{d t}

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