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

The integral of (3y-4y^3)^3 is (27y^4)/4-18y^6+18y^8-(32y^{10})/5+C

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integral of (3y-4y^3)^3

\int (3 y - 4 y^{3})^{3} d y

\frac{27 y ^{4}}{4} - 18 y^{6} + 18 y^{8} - \frac{32 y ^{10}}{5} + C

Solution steps

\int (3 y - 4 y^{3})^{3} d y

Expand

(3 y - 4 y^{3})^{3} : 27 y^{3} - 108 y^{5} + 144 y^{7} - 64 y^{9}

= \int 27 y^{3} - 108 y^{5} + 144 y^{7} - 64 y^{9} d y

Apply the Sum Rule:

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

= \int 27 y^{3} d y - \int 108 y^{5} d y + \int 144 y^{7} d y - \int 64 y^{9} d y

\int 27 y^{3} d y = \frac{27 y ^{4}}{4}

\int 108 y^{5} d y = 18 y^{6}

\int 144 y^{7} d y = 18 y^{8}

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

= \frac{27 y ^{4}}{4} - 18 y^{6} + 18 y^{8} - \frac{32 y ^{10}}{5}

Add a constant to the solution

= \frac{27 y ^{4}}{4} - 18 y^{6} + 18 y^{8} - \frac{32 y ^{10}}{5} + C

y = ln (\frac{x}{2}), at x = 4

\int \frac{t ^{2}}{t + 1} d t

\frac{d}{d x} (l x)

\frac{d}{d x} (arccos (- 3 x))

y^{^{''}} + 16 = 0

What is the integral of (3y-4y^3)^3 ?
The integral of (3y-4y^3)^3 is (27y^4)/4-18y^6+18y^8-(32y^{10})/5+C