What is the integral of (4x^3+1)^3(12x) ?

The integral of (4x^3+1)^3(12x) is 12((64x^{11}}{11}+6x^8+\frac{12x^5)/5+(x^2)/2)+C

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integral of (4x^3+1)^3(12x)

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

12 (\frac{64 x ^{11}}{11} + 6 x^{8} + \frac{12 x ^{5}}{5} + \frac{x ^{2}}{2}) + C

Solution steps

\int (4 x^{3} + 1)^{3} \cdot 12 x d x

Take the constant out:

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

= 12 \cdot \int (4 x^{3} + 1)^{3} x d x

Expand

(4 x^{3} + 1)^{3} x : 64 x^{10} + 48 x^{7} + 12 x^{4} + x

= 12 \cdot \int 64 x^{10} + 48 x^{7} + 12 x^{4} + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 12 (\int 64 x^{10} d x + \int 48 x^{7} d x + \int 12 x^{4} d x + \int x d x)

\int 64 x^{10} d x = \frac{64 x ^{11}}{11}

\int 48 x^{7} d x = 6 x^{8}

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

\int x d x = \frac{x ^{2}}{2}

= 12 (\frac{64 x ^{11}}{11} + 6 x^{8} + \frac{12 x ^{5}}{5} + \frac{x ^{2}}{2})

Add a constant to the solution

= 12 (\frac{64 x ^{11}}{11} + 6 x^{8} + \frac{12 x ^{5}}{5} + \frac{x ^{2}}{2}) + C

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What is the integral of (4x^3+1)^3(12x) ?
The integral of (4x^3+1)^3(12x) is 12((64x^{11}}{11}+6x^8+\frac{12x^5)/5+(x^2)/2)+C