What is the integral of sqrt(x)(x^{3/2}-4)^3 ?

The integral of sqrt(x)(x^{3/2}-4)^3 is 1/6 (x^{3/2}-4)^4+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)(x^{3/2}-4)^3

\int x (x^{\frac{3}{2}} - 4)^{3} d x

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

Solution steps

\int x (x^{\frac{3}{2}} - 4)^{3} d x

Apply u-substitution

: \int 2 u^{2} (u^{3} - 4)^{3} d u

= \int 2 u^{2} (u^{3} - 4)^{3} 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} - 4)^{3} d u

Apply u-substitution

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

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

Apply the Power Rule

: \frac{v ^{4}}{4}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int sin^{3} (x co) s^{18} x

y^{^{''}} - 3 y^{^{'}} + 2.25 y = 27 (x^{2} - x)

\int (6 x + 2) 3 x^{2} + 2 x d x

\frac{d}{d x} (4 x^{3} - x^{4})

\frac{d y}{d x} = k (10 - y) (5 - y)

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