What is the integral of 72(1+x^2)(7+18x+6x^3)^{16} ?

The integral of 72(1+x^2)(7+18x+6x^3)^{16} is 4/17 (6x^3+18x+7)^{17}+C

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integral of 72(1+x^2)(7+18x+6x^3)^{16}

\int 72 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{16} d x

\frac{4}{17} (6 x^{3} + 18 x + 7)^{17} + C

Solution steps

\int 72 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{16} d x

Take the constant out:

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

= 72 \cdot \int (1 + x^{2}) (7 + 18 x + 6 x^{3})^{16} d x

Apply u-substitution

: \int \frac{u ^{16}}{18} d u

= 72 \cdot \int \frac{u ^{16}}{18} d u

Take the constant out:

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

= 72 \cdot \frac{1}{18} \cdot \int u^{16} d u

Apply the Power Rule

: \frac{u ^{17}}{17}

= 72 \cdot \frac{1}{18} \cdot \frac{u ^{17}}{17}

Substitute back

u = 7 + 18 x + 6 x^{3}

= 72 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{17}}{17}

Simplify

72 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{17}}{17} : \frac{4}{17} (6 x^{3} + 18 x + 7)^{17}

= \frac{4}{17} (6 x^{3} + 18 x + 7)^{17}

Add a constant to the solution

= \frac{4}{17} (6 x^{3} + 18 x + 7)^{17} + C

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What is the integral of 72(1+x^2)(7+18x+6x^3)^{16} ?
The integral of 72(1+x^2)(7+18x+6x^3)^{16} is 4/17 (6x^3+18x+7)^{17}+C