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

y^{^{''}} + 2 \cdot 2.5 \cdot 4 \cdot y^{^{'}} + 16 y = 0

\int (x + \frac{1}{6 x}) d x

\int_{0}^{1} x (x^{2} - x) d x

\int \frac{x}{y} d y

n = 0 \sum \infty \frac{- 1 ^{n}}{n !}

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