What is the integral of (6x-5)sqrt((3x^2-5x)^3) ?

The integral of (6x-5)sqrt((3x^2-5x)^3) is 1/(720sqrt(3))(36x^2-60x)^{5/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x-5)sqrt((3x^2-5x)^3)

\int (6 x - 5) (3 x^{2} - 5 x)^{3} d x

\frac{1}{720 3} (36 x^{2} - 60 x)^{\frac{5}{2}} + C

Solution steps

\int (6 x - 5) (3 x^{2} - 5 x)^{3} d x

Apply u-substitution

: \int \frac{u ( u ^{2} - 25 ) ^{\frac{3}{2}}}{144 3} d u

= \int \frac{u ( u ^{2} - 25 ) ^{\frac{3}{2}}}{144 3} d u

Take the constant out:

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

= \frac{1}{144 3} \cdot \int u (u^{2} - 25)^{\frac{3}{2}} d u

Apply u-substitution

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

= \frac{1}{144 3} \cdot \int \frac{v ^{\frac{3}{2}}}{2} d v

Take the constant out:

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

= \frac{1}{144 3} \cdot \frac{1}{2} \cdot \int v^{\frac{3}{2}} d v

Apply the Power Rule

: \frac{2}{5} v^{\frac{5}{2}}

= \frac{1}{144 3} \cdot \frac{1}{2} \cdot \frac{2}{5} v^{\frac{5}{2}}

Substitute back

= \frac{1}{144 3} \cdot \frac{1}{2} \cdot \frac{2}{5} ((6 x - 5)^{2} - 25)^{\frac{5}{2}}

Simplify

\frac{1}{144 3} \cdot \frac{1}{2} \cdot \frac{2}{5} ((6 x - 5)^{2} - 25)^{\frac{5}{2}} : \frac{1}{720 3} (36 x^{2} - 60 x)^{\frac{5}{2}}

= \frac{1}{720 3} (36 x^{2} - 60 x)^{\frac{5}{2}}

Add a constant to the solution

= \frac{1}{720 3} (36 x^{2} - 60 x)^{\frac{5}{2}} + C

\int y^{\frac{3}{2}} d y

ln (cos (t))

\frac{d}{d x} ((x^{2} - 5 x)^{4})

n = 1 \sum \infty \frac{( ( - 1 ) ^{(n - 1)} )}{( ( 2 n - 1 ) ^{3} )}

\int \frac{x ^{2} - 4 x + 5}{( x ^{2} - 1 ) ( x + 3 )} d x

What is the integral of (6x-5)sqrt((3x^2-5x)^3) ?
The integral of (6x-5)sqrt((3x^2-5x)^3) is 1/(720sqrt(3))(36x^2-60x)^{5/2}+C