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

The integral of 36(1+x^2)(7+12x+4x^3)^{12} is 3/13 (4x^3+12x+7)^{13}+C

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

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

\frac{3}{13} (4 x^{3} + 12 x + 7)^{13} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

= 36 \cdot \frac{1}{12} \cdot \frac{u ^{13}}{13}

Substitute back

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

= 36 \cdot \frac{1}{12} \cdot \frac{( 7 + 12 x + 4 x ^{3} ) ^{13}}{13}

Simplify

36 \cdot \frac{1}{12} \cdot \frac{( 7 + 12 x + 4 x ^{3} ) ^{13}}{13} : \frac{3}{13} (4 x^{3} + 12 x + 7)^{13}

= \frac{3}{13} (4 x^{3} + 12 x + 7)^{13}

Add a constant to the solution

= \frac{3}{13} (4 x^{3} + 12 x + 7)^{13} + C

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What is the integral of 36(1+x^2)(7+12x+4x^3)^{12} ?
The integral of 36(1+x^2)(7+12x+4x^3)^{12} is 3/13 (4x^3+12x+7)^{13}+C