What is the integral of 60(1+x^2)(7+15x+5x^3)^{12} ?

The integral of 60(1+x^2)(7+15x+5x^3)^{12} is 4/13 (5x^3+15x+7)^{13}+C

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integral of 60(1+x^2)(7+15x+5x^3)^{12}

\int 60 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{12} d x

\frac{4}{13} (5 x^{3} + 15 x + 7)^{13} + C

Solution steps

\int 60 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{12} d x

Take the constant out:

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

= 60 \cdot \int (1 + x^{2}) (7 + 15 x + 5 x^{3})^{12} d x

Apply u-substitution

: \int \frac{u ^{12}}{15} d u

= 60 \cdot \int \frac{u ^{12}}{15} d u

Take the constant out:

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

= 60 \cdot \frac{1}{15} \cdot \int u^{12} d u

Apply the Power Rule

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

= 60 \cdot \frac{1}{15} \cdot \frac{u ^{13}}{13}

Substitute back

u = 7 + 15 x + 5 x^{3}

= 60 \cdot \frac{1}{15} \cdot \frac{( 7 + 15 x + 5 x ^{3} ) ^{13}}{13}

Simplify

60 \cdot \frac{1}{15} \cdot \frac{( 7 + 15 x + 5 x ^{3} ) ^{13}}{13} : \frac{4}{13} (5 x^{3} + 15 x + 7)^{13}

= \frac{4}{13} (5 x^{3} + 15 x + 7)^{13}

Add a constant to the solution

= \frac{4}{13} (5 x^{3} + 15 x + 7)^{13} + C

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What is the integral of 60(1+x^2)(7+15x+5x^3)^{12} ?
The integral of 60(1+x^2)(7+15x+5x^3)^{12} is 4/13 (5x^3+15x+7)^{13}+C