What is the integral of (-16-6x^4)/(x^2) ?

The integral of (-16-6x^4)/(x^2) is (16)/x-2x^3+C

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integral of (-16-6x^4)/(x^2)

integral

\frac{- 16 - 6 x ^{4}}{x ^{2}}

\frac{16}{x} - 2 x^{3} + C

Solution steps

\int \frac{- 16 - 6 x ^{4}}{x ^{2}} d x

Expand

\frac{- 16 - 6 x ^{4}}{x ^{2}} : - \frac{16}{x ^{2}} - 6 x^{2}

= \int - \frac{16}{x ^{2}} - 6 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 \frac{16}{x ^{2}} d x - \int 6 x^{2} d x

\int \frac{16}{x ^{2}} d x = - \frac{16}{x}

\int 6 x^{2} d x = 2 x^{3}

= - (- \frac{16}{x}) - 2 x^{3}

Simplify

= \frac{16}{x} - 2 x^{3}

Add a constant to the solution

= \frac{16}{x} - 2 x^{3} + C

\frac{1}{( 1 + 4 \frac{x}{l} )}

\frac{2 x}{x ^{2}}

\frac{x ^{4} + x ^{2} + 1}{( x + 1 )}

\frac{2 x}{( x ^{2} + 1 )}

e^{x^{3}}

What is the integral of (-16-6x^4)/(x^2) ?
The integral of (-16-6x^4)/(x^2) is (16)/x-2x^3+C