What is the integral of z^2(z^3-1)^6 ?

The integral of z^2(z^3-1)^6 is -1/21 (-z^3+1)^7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of z^2(z^3-1)^6

\int z^{2} (z^{3} - 1)^{6} d z

- \frac{1}{21} (- z^{3} + 1)^{7} + C

Solution steps

\int z^{2} (z^{3} - 1)^{6} d z

Factor

z^{3} - 1 : - (- z^{3} + 1)

= \int z^{2} (- (- z^{3} + 1))^{6} d z

(- (- z^{3} + 1))^{6} = (- z^{3} + 1)^{6}

= \int z^{2} (- z^{3} + 1)^{6} d z

Apply u-substitution

: \int \frac{u ( - u ^{\frac{3}{2}} + 1 ) ^{6}}{2 z} d u

= \int \frac{( - u + 1 ) ^{6}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (- u + 1)^{6} d u

Apply u-substitution

: \int - v^{6} d v

= \frac{1}{3} \cdot \int - v^{6} d v

Take the constant out:

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

= \frac{1}{3} (- \int v^{6} d v)

Apply the Power Rule

: \frac{v ^{7}}{7}

= \frac{1}{3} (- \frac{v ^{7}}{7})

Substitute back

= \frac{1}{3} (- \frac{( - z ^{3} + 1 ) ^{7}}{7})

Simplify

\frac{1}{3} (- \frac{( - z ^{3} + 1 ) ^{7}}{7}) : - \frac{1}{21} (- z^{3} + 1)^{7}

= - \frac{1}{21} (- z^{3} + 1)^{7}

Add a constant to the solution

= - \frac{1}{21} (- z^{3} + 1)^{7} + C

\int p (p + 6)^{5} d p

\int \frac{2}{1 - 4 x ^{2}} d x

f (x) = 2 x - x^{2}, at x = 1

\frac{d}{d x} (\frac{x ^{3} - 1}{ln ( x )})

\int cos (x) sin (x) 1 + cos^{2} (x) d x

What is the integral of z^2(z^3-1)^6 ?
The integral of z^2(z^3-1)^6 is -1/21 (-z^3+1)^7+C