What is the integral of (3x^2)/(2sqrt(5-3x^3)) ?

The integral of (3x^2)/(2sqrt(5-3x^3)) is -1/3 sqrt(5-3x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2)/(2sqrt(5-3x^3))

\int \frac{3 x ^{2}}{2 5 - 3 x ^{3}} d x

- \frac{1}{3} 5 - 3 x^{3} + C

Solution steps

\int \frac{3 x ^{2}}{2 5 - 3 x ^{3}} d x

Take the constant out:

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

= \frac{3}{2} \cdot \int \frac{x ^{2}}{5 - 3 x ^{3}} d x

Apply u-substitution

: \int \frac{u}{2 x 5 - 3 u ^{\frac{3}{2}}} d u

= \frac{3}{2} \cdot \int \frac{1}{3 5 - 3 u} d u

Take the constant out:

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

= \frac{3}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{5 - 3 u} d u

Apply u-substitution

: \int - \frac{1}{3 v} d v

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

Take the constant out:

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

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

Apply the Power Rule

: 2 v^{\frac{1}{2}}

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

Substitute back

= \frac{3}{2} \cdot \frac{1}{3} (- \frac{1}{3} \cdot 2 (5 - 3 x^{3})^{\frac{1}{2}})

Simplify

\frac{3}{2} \cdot \frac{1}{3} (- \frac{1}{3} \cdot 2 (5 - 3 x^{3})^{\frac{1}{2}}) : - \frac{1}{3} 5 - 3 x^{3}

= - \frac{1}{3} 5 - 3 x^{3}

Add a constant to the solution

= - \frac{1}{3} 5 - 3 x^{3} + C

\int (y + \frac{1}{y}) d y

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

\int \frac{4 x}{2 x ^{2} + 6} d x

\int 2 - cos (x) sin (x) d x

\int sec^{2} (θ) sin^{3} (θ) d θ

What is the integral of (3x^2)/(2sqrt(5-3x^3)) ?
The integral of (3x^2)/(2sqrt(5-3x^3)) is -1/3 sqrt(5-3x^3)+C