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

The integral of sqrt(x)sqrt(x^{3/2)-1} is 4/9 (x^{3/2}-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 sqrt(x)sqrt(x^{3/2)-1}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{v}{3} d v

= 2 \cdot \int \frac{v}{3} d v

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (lo g_{5} (2 x^{2} + 1))

(2 x + \frac{2}{y}) + (\frac{( 2 x ^{2} )}{y} + \frac{2 x}{y ^{2}}) y^{^{'}} = 0

f (x) = x^{4} - 10 x^{2} + 9

\int - \frac{14}{x ^{15}} d x

\int x^{n} d x

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