What is the integral of ((3x^5-4)^2*5x^4) ?

The integral of ((3x^5-4)^2*5x^4) is 3x^{15}-12x^{10}+16x^5+C

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integral of ((3x^5-4)^2*5x^4)

\int ((3 x^{5} - 4)^{2} \cdot 5 x^{4}) d x

3 x^{15} - 12 x^{10} + 16 x^{5} + C

Solution steps

\int (3 x^{5} - 4)^{2} \cdot 5 x^{4} d x

Take the constant out:

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

= 5 \cdot \int (3 x^{5} - 4)^{2} x^{4} d x

Expand

(3 x^{5} - 4)^{2} x^{4} : 9 x^{14} - 24 x^{9} + 16 x^{4}

= 5 \cdot \int 9 x^{14} - 24 x^{9} + 16 x^{4} 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

= 5 (\int 9 x^{14} d x - \int 24 x^{9} d x + \int 16 x^{4} d x)

\int 9 x^{14} d x = \frac{3 x ^{15}}{5}

\int 24 x^{9} d x = \frac{12 x ^{10}}{5}

\int 16 x^{4} d x = \frac{16 x ^{5}}{5}

= 5 (\frac{3 x ^{15}}{5} - \frac{12 x ^{10}}{5} + \frac{16 x ^{5}}{5})

Simplify

5 (\frac{3 x ^{15}}{5} - \frac{12 x ^{10}}{5} + \frac{16 x ^{5}}{5}) : 3 x^{15} - 12 x^{10} + 16 x^{5}

= 3 x^{15} - 12 x^{10} + 16 x^{5}

Add a constant to the solution

= 3 x^{15} - 12 x^{10} + 16 x^{5} + C

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What is the integral of ((3x^5-4)^2*5x^4) ?
The integral of ((3x^5-4)^2*5x^4) is 3x^{15}-12x^{10}+16x^5+C