What is the integral of 16-8x^2+x^4 ?

The integral of 16-8x^2+x^4 is 16x-(8x^3)/3+(x^5)/5+C

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integral of 16-8x^2+x^4

\int 16 - 8 x^{2} + x^{4} d x

16 x - \frac{8 x ^{3}}{3} + \frac{x ^{5}}{5} + C

Solution steps

\int 16 - 8 x^{2} + 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

= \int 16 d x - \int 8 x^{2} d x + \int x^{4} d x

\int 16 d x = 16 x

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

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

= 16 x - \frac{8 x ^{3}}{3} + \frac{x ^{5}}{5}

Add a constant to the solution

= 16 x - \frac{8 x ^{3}}{3} + \frac{x ^{5}}{5} + C

\int a x^{2} d x

f (x) = \frac{tan ( x ) - 1}{sec ( x )}

2 x^{2} y^{^{'}} = π y sin (\frac{π}{2 x})

x^{6 c o s (x)}

\frac{d}{d x} (8^{x^{4}})

What is the integral of 16-8x^2+x^4 ?
The integral of 16-8x^2+x^4 is 16x-(8x^3)/3+(x^5)/5+C