What is the integral of 2x(4x-3)^3 ?

The integral of 2x(4x-3)^3 is (128x^5)/5-72x^4+72x^3-27x^2+C

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integral of 2x(4x-3)^3

\int 2 x (4 x - 3)^{3} d x

\frac{128 x ^{5}}{5} - 72 x^{4} + 72 x^{3} - 27 x^{2} + C

Solution steps

\int 2 x (4 x - 3)^{3} d x

Take the constant out:

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

= 2 \cdot \int x (4 x - 3)^{3} d x

Expand

x (4 x - 3)^{3} : 64 x^{4} - 144 x^{3} + 108 x^{2} - 27 x

= 2 \cdot \int 64 x^{4} - 144 x^{3} + 108 x^{2} - 27 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

= 2 (\int 64 x^{4} d x - \int 144 x^{3} d x + \int 108 x^{2} d x - \int 27 x d x)

\int 64 x^{4} d x = \frac{64 x ^{5}}{5}

\int 144 x^{3} d x = 36 x^{4}

\int 108 x^{2} d x = 36 x^{3}

\int 27 x d x = \frac{27 x ^{2}}{2}

= 2 (\frac{64 x ^{5}}{5} - 36 x^{4} + 36 x^{3} - \frac{27 x ^{2}}{2})

Simplify

2 (\frac{64 x ^{5}}{5} - 36 x^{4} + 36 x^{3} - \frac{27 x ^{2}}{2}) : \frac{128 x ^{5}}{5} - 72 x^{4} + 72 x^{3} - 27 x^{2}

= \frac{128 x ^{5}}{5} - 72 x^{4} + 72 x^{3} - 27 x^{2}

Add a constant to the solution

= \frac{128 x ^{5}}{5} - 72 x^{4} + 72 x^{3} - 27 x^{2} + C

d er i v a t i v e \frac{2}{( 3 x ^{2} + 7 ) ^{2}}

s l o p e f (x) = 9 x

\int (sin (x))^{3} d x

\frac{\partial}{\partial r} (r)

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

What is the integral of 2x(4x-3)^3 ?
The integral of 2x(4x-3)^3 is (128x^5)/5-72x^4+72x^3-27x^2+C