What is the integral of 64x^3(4x^2+1)^3 ?

The integral of 64x^3(4x^2+1)^3 is 64((32x^{10})/5+6x^8+2x^6+(x^4)/4)+C

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integral of 64x^3(4x^2+1)^3

\int 64 x^{3} (4 x^{2} + 1)^{3} d x

64 (\frac{32 x ^{10}}{5} + 6 x^{8} + 2 x^{6} + \frac{x ^{4}}{4}) + C

Solution steps

\int 64 x^{3} (4 x^{2} + 1)^{3} d x

Take the constant out:

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

= 64 \cdot \int x^{3} (4 x^{2} + 1)^{3} d x

Expand

x^{3} (4 x^{2} + 1)^{3} : 64 x^{9} + 48 x^{7} + 12 x^{5} + x^{3}

= 64 \cdot \int 64 x^{9} + 48 x^{7} + 12 x^{5} + x^{3} 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

= 64 (\int 64 x^{9} d x + \int 48 x^{7} d x + \int 12 x^{5} d x + \int x^{3} d x)

\int 64 x^{9} d x = \frac{32 x ^{10}}{5}

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

\int 12 x^{5} d x = 2 x^{6}

\int x^{3} d x = \frac{x ^{4}}{4}

= 64 (\frac{32 x ^{10}}{5} + 6 x^{8} + 2 x^{6} + \frac{x ^{4}}{4})

Add a constant to the solution

= 64 (\frac{32 x ^{10}}{5} + 6 x^{8} + 2 x^{6} + \frac{x ^{4}}{4}) + C

\int x^{2} (x^{3} + 2)^{5} d x

\frac{d}{d x} (\frac{cot ^{2} ( 2 x )}{1 + x ^{2}})

2 x (x + 5)

\frac{d}{d t} (6 cos^{3} (t))

\int \frac{( 3 x ^{3} - 5 x ^{2} + 10 x - 3 )}{3 x + 1} d x

What is the integral of 64x^3(4x^2+1)^3 ?
The integral of 64x^3(4x^2+1)^3 is 64((32x^{10})/5+6x^8+2x^6+(x^4)/4)+C