What is the integral of 16z^2(64z^3+1)^{12} ?

The integral of 16z^2(64z^3+1)^{12} is 1/156 (64z^3+1)^{13}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 16z^2(64z^3+1)^{12}

\int 16 z^{2} (64 z^{3} + 1)^{12} d z

\frac{1}{156} (64 z^{3} + 1)^{13} + C

Solution steps

\int 16 z^{2} (64 z^{3} + 1)^{12} d z

Take the constant out:

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

= 16 \cdot \int z^{2} (64 z^{3} + 1)^{12} d z

Apply u-substitution

: \int \frac{u ( 64 u ^{\frac{3}{2}} + 1 ) ^{12}}{2 z} d u

= 16 \cdot \int \frac{( 64 u + 1 ) ^{12}}{3} d u

Take the constant out:

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

= 16 \cdot \frac{1}{3} \cdot \int (64 u + 1)^{12} d u

Apply u-substitution

: \int \frac{v ^{12}}{64} d v

= 16 \cdot \frac{1}{3} \cdot \int \frac{v ^{12}}{64} d v

Take the constant out:

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

= 16 \cdot \frac{1}{3} \cdot \frac{1}{64} \cdot \int v^{12} d v

Apply the Power Rule

: \frac{v ^{13}}{13}

= 16 \cdot \frac{1}{3} \cdot \frac{1}{64} \cdot \frac{v ^{13}}{13}

Substitute back

= 16 \cdot \frac{1}{3} \cdot \frac{1}{64} \cdot \frac{( 64 z ^{3} + 1 ) ^{13}}{13}

Simplify

16 \cdot \frac{1}{3} \cdot \frac{1}{64} \cdot \frac{( 64 z ^{3} + 1 ) ^{13}}{13} : \frac{1}{156} (64 z^{3} + 1)^{13}

= \frac{1}{156} (64 z^{3} + 1)^{13}

Add a constant to the solution

= \frac{1}{156} (64 z^{3} + 1)^{13} + C

\int sin (θ^{2}) d θ

\int p (p + 7)^{6} d p

\int sin (x) cos (4 x) d x

\int (\frac{arctan ( x )}{x x}) d x

\int [1 - cos^{2} (θ) sec (θ)] d θ

What is the integral of 16z^2(64z^3+1)^{12} ?
The integral of 16z^2(64z^3+1)^{12} is 1/156 (64z^3+1)^{13}+C