What is the integral of (x^7+3x^5-2x^3-4)/(x^3) ?

The integral of (x^7+3x^5-2x^3-4)/(x^3) is (x^5)/5+x^3-2x+2/(x^2)+C

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integral of (x^7+3x^5-2x^3-4)/(x^3)

\int \frac{x ^{7} + 3 x ^{5} - 2 x ^{3} - 4}{x ^{3}} d x

\frac{x ^{5}}{5} + x^{3} - 2 x + \frac{2}{x ^{2}} + C

Solution steps

\int \frac{x ^{7} + 3 x ^{5} - 2 x ^{3} - 4}{x ^{3}} d x

Expand

\frac{x ^{7} + 3 x ^{5} - 2 x ^{3} - 4}{x ^{3}} : x^{4} + 3 x^{2} - 2 - \frac{4}{x ^{3}}

= \int x^{4} + 3 x^{2} - 2 - \frac{4}{x ^{3}} 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 x^{4} d x + \int 3 x^{2} d x - \int 2 d x - \int \frac{4}{x ^{3}} d x

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

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

\int 2 d x = 2 x

\int \frac{4}{x ^{3}} d x = - \frac{2}{x ^{2}}

= \frac{x ^{5}}{5} + x^{3} - 2 x - (- \frac{2}{x ^{2}})

Simplify

= \frac{x ^{5}}{5} + x^{3} - 2 x + \frac{2}{x ^{2}}

Add a constant to the solution

= \frac{x ^{5}}{5} + x^{3} - 2 x + \frac{2}{x ^{2}} + C

\frac{d}{d x} (\frac{e ^{1 - c o s (x)}}{cos ( x )})

\frac{d}{d u} (cos (c) (cot (u)))

y^{^{'}} = e^{2 x + y}

\int \frac{5}{5 + e ^{x}} d x

\frac{\partial}{\partial x} (ln (1 - e^{- x}))

What is the integral of (x^7+3x^5-2x^3-4)/(x^3) ?
The integral of (x^7+3x^5-2x^3-4)/(x^3) is (x^5)/5+x^3-2x+2/(x^2)+C