What is the integral of x^2(6x^3-7)^2 ?

The integral of x^2(6x^3-7)^2 is 1/3 (12x^9-42x^6+49x^3)+C

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integral of x^2(6x^3-7)^2

\int x^{2} (6 x^{3} - 7)^{2} d x

\frac{1}{3} (12 x^{9} - 42 x^{6} + 49 x^{3}) + C

Solution steps

\int x^{2} (6 x^{3} - 7)^{2} d x

Apply u-substitution

: \int \frac{u ( 6 u ^{\frac{3}{2}} - 7 ) ^{2}}{2 x} d u

= \int \frac{( 6 u - 7 ) ^{2}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (6 u - 7)^{2} d u

Expand

(6 u - 7)^{2} : 36 u^{2} - 84 u + 49

= \frac{1}{3} \cdot \int 36 u^{2} - 84 u + 49 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}{3} (\int 36 u^{2} d u - \int 84 u d u + \int 49 d u)

\int 36 u^{2} d u = 12 u^{3}

\int 84 u d u = 42 u^{2}

\int 49 d u = 49 u

= \frac{1}{3} (12 u^{3} - 42 u^{2} + 49 u)

Substitute back

u = x^{3}

= \frac{1}{3} (12 (x^{3})^{3} - 42 (x^{3})^{2} + 49 x^{3})

Simplify

\frac{1}{3} (12 (x^{3})^{3} - 42 (x^{3})^{2} + 49 x^{3}) : \frac{1}{3} (12 x^{9} - 42 x^{6} + 49 x^{3})

= \frac{1}{3} (12 x^{9} - 42 x^{6} + 49 x^{3})

Add a constant to the solution

= \frac{1}{3} (12 x^{9} - 42 x^{6} + 49 x^{3}) + C

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What is the integral of x^2(6x^3-7)^2 ?
The integral of x^2(6x^3-7)^2 is 1/3 (12x^9-42x^6+49x^3)+C