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

The integral of (x^4-3x^2+4x)/(5x) is 1/5 ((x^4)/4-(3x^2)/2+4x)+C

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

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

\frac{1}{5} (\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + 4 x) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{x ^{4} - 3 x ^{2} + 4 x}{x} d x

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} (\int \frac{x ^{4}}{x} d x - \int \frac{3 x ^{2}}{x} d x + \int \frac{4 x}{x} d x)

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

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

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

= \frac{1}{5} (\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + 4 x)

Add a constant to the solution

= \frac{1}{5} (\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + 4 x) + C

x \to \infty lim (\frac{e ^{x}}{1 - e ^{- x}})

x \to 1 + lim (\frac{3}{x ^{3} - 1})

- 9 e^{x^{- 9}}

\int \frac{1}{x ( ln ( 2 x ) )} d x

(40 cos (20 t - \frac{π}{2}))^{^{'}}

What is the integral of (x^4-3x^2+4x)/(5x) ?
The integral of (x^4-3x^2+4x)/(5x) is 1/5 ((x^4)/4-(3x^2)/2+4x)+C