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

The integral of (x^5-4x)^2(5x^4-4) is (x^{15})/3-4x^{11}+16x^7-(64x^3)/3+C

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

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

\frac{x ^{15}}{3} - 4 x^{11} + 16 x^{7} - \frac{64 x ^{3}}{3} + C

Solution steps

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

Expand

(x^{5} - 4 x)^{2} (5 x^{4} - 4) : 5 x^{14} - 44 x^{10} + 112 x^{6} - 64 x^{2}

= \int 5 x^{14} - 44 x^{10} + 112 x^{6} - 64 x^{2} 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

= \int 5 x^{14} d x - \int 44 x^{10} d x + \int 112 x^{6} d x - \int 64 x^{2} d x

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

\int 44 x^{10} d x = 4 x^{11}

\int 112 x^{6} d x = 16 x^{7}

\int 64 x^{2} d x = \frac{64 x ^{3}}{3}

= \frac{x ^{15}}{3} - 4 x^{11} + 16 x^{7} - \frac{64 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{15}}{3} - 4 x^{11} + 16 x^{7} - \frac{64 x ^{3}}{3} + C

x \to 0 + lim (\frac{1}{1 + \frac{1}{x}})

\frac{d}{d x} (- 8)

x \to 0 lim (\frac{sin ^{2} ( 3 x )}{5 x ^{2}})

\int sin (4 x) sin (x) d x

2 (t + 2)

What is the integral of (x^5-4x)^2(5x^4-4) ?
The integral of (x^5-4x)^2(5x^4-4) is (x^{15})/3-4x^{11}+16x^7-(64x^3)/3+C