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

The integral of (3x^2-4)^2x^3 is 1/2 ((9x^8)/4-8x^6+8x^4)+C

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

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

\frac{1}{2} (\frac{9 x ^{8}}{4} - 8 x^{6} + 8 x^{4}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( 3 u - 4 ) ^{2}}{2} d u

= \int \frac{u ( 3 u - 4 ) ^{2}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u (3 u - 4)^{2} d u

Factor

3 u - 4 : - (- 3 u + 4)

= \frac{1}{2} \cdot \int u (- (- 3 u + 4))^{2} d u

(- (- 3 u + 4))^{2} = (- 3 u + 4)^{2}

= \frac{1}{2} \cdot \int u (- 3 u + 4)^{2} d u

Expand

u (- 3 u + 4)^{2} : 9 u^{3} - 24 u^{2} + 16 u

= \frac{1}{2} \cdot \int 9 u^{3} - 24 u^{2} + 16 u d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int 9 u^{3} d u - \int 24 u^{2} d u + \int 16 u d u)

\int 9 u^{3} d u = \frac{9 u ^{4}}{4}

\int 24 u^{2} d u = 8 u^{3}

\int 16 u d u = 8 u^{2}

= \frac{1}{2} (\frac{9 u ^{4}}{4} - 8 u^{3} + 8 u^{2})

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{9 ( x ^{2} ) ^{4}}{4} - 8 (x^{2})^{3} + 8 (x^{2})^{2})

Simplify

\frac{1}{2} (\frac{9 ( x ^{2} ) ^{4}}{4} - 8 (x^{2})^{3} + 8 (x^{2})^{2}) : \frac{1}{2} (\frac{9 x ^{8}}{4} - 8 x^{6} + 8 x^{4})

= \frac{1}{2} (\frac{9 x ^{8}}{4} - 8 x^{6} + 8 x^{4})

Add a constant to the solution

= \frac{1}{2} (\frac{9 x ^{8}}{4} - 8 x^{6} + 8 x^{4}) + C

\frac{\partial}{\partial y} (e^{x})

\frac{\partial}{\partial x} (5 y - 2 x)

30 \frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} = 0

x \to 0 lim (\frac{x + x ^{2}}{∣ x ∣})

\frac{d}{d x} (x (x^{2} - 5))

What is the integral of (3x^2-4)^2x^3 ?
The integral of (3x^2-4)^2x^3 is 1/2 ((9x^8)/4-8x^6+8x^4)+C