What is the integral of (7x^2-4)^2x^3 ?

The integral of (7x^2-4)^2x^3 is 1/2 ((49x^8)/4-(56x^6)/3+8x^4)+C

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integral of (7x^2-4)^2x^3

\int (7 x^{2} - 4)^{2} x^{3} d x

\frac{1}{2} (\frac{49 x ^{8}}{4} - \frac{56 x ^{6}}{3} + 8 x^{4}) + C

Solution steps

\int (7 x^{2} - 4)^{2} x^{3} d x

Apply u-substitution

: \int \frac{u ( 7 u - 4 ) ^{2}}{2} d u

= \int \frac{u ( 7 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 (7 u - 4)^{2} d u

Factor

7 u - 4 : - (- 7 u + 4)

= \frac{1}{2} \cdot \int u (- (- 7 u + 4))^{2} d u

(- (- 7 u + 4))^{2} = (- 7 u + 4)^{2}

= \frac{1}{2} \cdot \int u (- 7 u + 4)^{2} d u

Expand

u (- 7 u + 4)^{2} : 49 u^{3} - 56 u^{2} + 16 u

= \frac{1}{2} \cdot \int 49 u^{3} - 56 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 49 u^{3} d u - \int 56 u^{2} d u + \int 16 u d u)

\int 49 u^{3} d u = \frac{49 u ^{4}}{4}

\int 56 u^{2} d u = \frac{56 u ^{3}}{3}

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

= \frac{1}{2} (\frac{49 u ^{4}}{4} - \frac{56 u ^{3}}{3} + 8 u^{2})

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{49 ( x ^{2} ) ^{4}}{4} - \frac{56 ( x ^{2} ) ^{3}}{3} + 8 (x^{2})^{2})

Simplify

\frac{1}{2} (\frac{49 ( x ^{2} ) ^{4}}{4} - \frac{56 ( x ^{2} ) ^{3}}{3} + 8 (x^{2})^{2}) : \frac{1}{2} (\frac{49 x ^{8}}{4} - \frac{56 x ^{6}}{3} + 8 x^{4})

= \frac{1}{2} (\frac{49 x ^{8}}{4} - \frac{56 x ^{6}}{3} + 8 x^{4})

Add a constant to the solution

= \frac{1}{2} (\frac{49 x ^{8}}{4} - \frac{56 x ^{6}}{3} + 8 x^{4}) + C

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What is the integral of (7x^2-4)^2x^3 ?
The integral of (7x^2-4)^2x^3 is 1/2 ((49x^8)/4-(56x^6)/3+8x^4)+C