What is the integral of 49z^2(343z^3+1)^{12} ?

The integral of 49z^2(343z^3+1)^{12} is 1/273 (343z^3+1)^{13}+C

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integral of 49z^2(343z^3+1)^{12}

\int 49 z^{2} (343 z^{3} + 1)^{12} d z

\frac{1}{273} (343 z^{3} + 1)^{13} + C

Solution steps

\int 49 z^{2} (343 z^{3} + 1)^{12} d z

Take the constant out:

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

= 49 \cdot \int z^{2} (343 z^{3} + 1)^{12} d z

Apply u-substitution

: \int \frac{u ( 343 u ^{\frac{3}{2}} + 1 ) ^{12}}{2 z} d u

= 49 \cdot \int \frac{( 343 u + 1 ) ^{12}}{3} d u

Take the constant out:

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

= 49 \cdot \frac{1}{3} \cdot \int (343 u + 1)^{12} d u

Apply u-substitution

: \int \frac{v ^{12}}{343} d v

= 49 \cdot \frac{1}{3} \cdot \int \frac{v ^{12}}{343} d v

Take the constant out:

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

= 49 \cdot \frac{1}{3} \cdot \frac{1}{343} \cdot \int v^{12} d v

Apply the Power Rule

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

= 49 \cdot \frac{1}{3} \cdot \frac{1}{343} \cdot \frac{v ^{13}}{13}

Substitute back

= 49 \cdot \frac{1}{3} \cdot \frac{1}{343} \cdot \frac{( 343 z ^{3} + 1 ) ^{13}}{13}

Simplify

49 \cdot \frac{1}{3} \cdot \frac{1}{343} \cdot \frac{( 343 z ^{3} + 1 ) ^{13}}{13} : \frac{1}{273} (343 z^{3} + 1)^{13}

= \frac{1}{273} (343 z^{3} + 1)^{13}

Add a constant to the solution

= \frac{1}{273} (343 z^{3} + 1)^{13} + C

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What is the integral of 49z^2(343z^3+1)^{12} ?
The integral of 49z^2(343z^3+1)^{12} is 1/273 (343z^3+1)^{13}+C