What is the integral of 36(1+x^2)(7+12x+4x^3)^{16} ?

The integral of 36(1+x^2)(7+12x+4x^3)^{16} is 3/17 (4x^3+12x+7)^{17}+C

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integral of 36(1+x^2)(7+12x+4x^3)^{16}

\int 36 (1 + x^{2}) (7 + 12 x + 4 x^{3})^{16} d x

\frac{3}{17} (4 x^{3} + 12 x + 7)^{17} + C

Solution steps

\int 36 (1 + x^{2}) (7 + 12 x + 4 x^{3})^{16} 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 + 12 x + 4 x^{3})^{16} d x

Apply u-substitution

: \int \frac{u ^{16}}{12} d u

= 36 \cdot \int \frac{u ^{16}}{12} d u

Take the constant out:

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

= 36 \cdot \frac{1}{12} \cdot \int u^{16} d u

Apply the Power Rule

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

= 36 \cdot \frac{1}{12} \cdot \frac{u ^{17}}{17}

Substitute back

u = 7 + 12 x + 4 x^{3}

= 36 \cdot \frac{1}{12} \cdot \frac{( 7 + 12 x + 4 x ^{3} ) ^{17}}{17}

Simplify

36 \cdot \frac{1}{12} \cdot \frac{( 7 + 12 x + 4 x ^{3} ) ^{17}}{17} : \frac{3}{17} (4 x^{3} + 12 x + 7)^{17}

= \frac{3}{17} (4 x^{3} + 12 x + 7)^{17}

Add a constant to the solution

= \frac{3}{17} (4 x^{3} + 12 x + 7)^{17} + C

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What is the integral of 36(1+x^2)(7+12x+4x^3)^{16} ?
The integral of 36(1+x^2)(7+12x+4x^3)^{16} is 3/17 (4x^3+12x+7)^{17}+C