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

The integral of-4x(-2x^2+2)^5 is -4(-(8x^{12})/3+16x^{10}-40x^8+(160x^6)/3-40x^4+16x^2)+C

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

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

- 4 (- \frac{8 x ^{12}}{3} + 16 x^{10} - 40 x^{8} + \frac{160 x ^{6}}{3} - 40 x^{4} + 16 x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Expand

x (- 2 x^{2} + 2)^{5} : - 32 x^{11} + 160 x^{9} - 320 x^{7} + 320 x^{5} - 160 x^{3} + 32 x

= - 4 \cdot \int - 32 x^{11} + 160 x^{9} - 320 x^{7} + 320 x^{5} - 160 x^{3} + 32 x 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

= - 4 (- \int 32 x^{11} d x + \int 160 x^{9} d x - \int 320 x^{7} d x + \int 320 x^{5} d x - \int 160 x^{3} d x + \int 32 x d x)

\int 32 x^{11} d x = \frac{8 x ^{12}}{3}

\int 160 x^{9} d x = 16 x^{10}

\int 320 x^{7} d x = 40 x^{8}

\int 320 x^{5} d x = \frac{160 x ^{6}}{3}

\int 160 x^{3} d x = 40 x^{4}

\int 32 x d x = 16 x^{2}

= - 4 (- \frac{8 x ^{12}}{3} + 16 x^{10} - 40 x^{8} + \frac{160 x ^{6}}{3} - 40 x^{4} + 16 x^{2})

Add a constant to the solution

= - 4 (- \frac{8 x ^{12}}{3} + 16 x^{10} - 40 x^{8} + \frac{160 x ^{6}}{3} - 40 x^{4} + 16 x^{2}) + C

y = \frac{e ^{8 x}}{e ^{8 x} + 1}

\frac{50}{100 + 0.01 s}

x \to 0 lim (\frac{1}{( x + 1 ) ^{2}})

(e^{x^{2} + 1})^{^{'}}

x^{3} os (x)

What is the integral of-4x(-2x^2+2)^5 ?
The integral of-4x(-2x^2+2)^5 is -4(-(8x^{12})/3+16x^{10}-40x^8+(160x^6)/3-40x^4+16x^2)+C