What is the integral of (4x+9)^{1/2} ?

The integral of (4x+9)^{1/2} is 1/6 (4x+9)^{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 (4x+9)^{1/2}

integral

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

\frac{1}{6} (4 x + 9)^{\frac{3}{2}} + C

Solution steps

\int (4 x + 9)^{\frac{1}{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 4 x + 9

= \frac{1}{4} \cdot \frac{2}{3} (4 x + 9)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (4 x + 9)^{\frac{3}{2}} : \frac{1}{6} (4 x + 9)^{\frac{3}{2}}

= \frac{1}{6} (4 x + 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (4 x + 9)^{\frac{3}{2}} + C

(2 x + 8)^{3}

x^{2} arcsin (x)

\frac{1}{1 + x ^{3}}

\frac{( 1 + x )}{( 1 - x )}

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

What is the integral of (4x+9)^{1/2} ?
The integral of (4x+9)^{1/2} is 1/6 (4x+9)^{3/2}+C