What is the integral of 36(1+x^2)(7+18x+6x^3)^{12} ?

The integral of 36(1+x^2)(7+18x+6x^3)^{12} is 2/13 (6x^3+18x+7)^{13}+C

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integral of 36(1+x^2)(7+18x+6x^3)^{12}

\int 36 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{12} d x

\frac{2}{13} (6 x^{3} + 18 x + 7)^{13} + C

Solution steps

\int 36 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{12} d x

Take the constant out:

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

= 36 \cdot \int (1 + x^{2}) (7 + 18 x + 6 x^{3})^{12} d x

Apply u-substitution

: \int \frac{u ^{12}}{18} d u

= 36 \cdot \int \frac{u ^{12}}{18} d u

Take the constant out:

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

= 36 \cdot \frac{1}{18} \cdot \int u^{12} d u

Apply the Power Rule

: \frac{u ^{13}}{13}

= 36 \cdot \frac{1}{18} \cdot \frac{u ^{13}}{13}

Substitute back

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

= 36 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{13}}{13}

Simplify

36 \cdot \frac{1}{18} \cdot \frac{( 7 + 18 x + 6 x ^{3} ) ^{13}}{13} : \frac{2}{13} (6 x^{3} + 18 x + 7)^{13}

= \frac{2}{13} (6 x^{3} + 18 x + 7)^{13}

Add a constant to the solution

= \frac{2}{13} (6 x^{3} + 18 x + 7)^{13} + C

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What is the integral of 36(1+x^2)(7+18x+6x^3)^{12} ?
The integral of 36(1+x^2)(7+18x+6x^3)^{12} is 2/13 (6x^3+18x+7)^{13}+C