What is the integral of (15x^4-6x^3+5)/(5x^2) ?

The integral of (15x^4-6x^3+5)/(5x^2) is x^3-3/5 x^2-1/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (15x^4-6x^3+5)/(5x^2)

\int \frac{15 x ^{4} - 6 x ^{3} + 5}{5 x ^{2}} d x

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

Solution steps

\int \frac{15 x ^{4} - 6 x ^{3} + 5}{5 x ^{2}} 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{15 x ^{4} - 6 x ^{3} + 5}{x ^{2}} d x

Expand

\frac{15 x ^{4} - 6 x ^{3} + 5}{x ^{2}} : 15 x^{2} - 6 x + \frac{5}{x ^{2}}

= \frac{1}{5} \cdot \int 15 x^{2} - 6 x + \frac{5}{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

= \frac{1}{5} (\int 15 x^{2} d x - \int 6 x d x + \int \frac{5}{x ^{2}} d x)

\int 15 x^{2} d x = 5 x^{3}

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

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

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

Simplify

\frac{1}{5} (5 x^{3} - 3 x^{2} - \frac{5}{x}) : x^{3} - \frac{3}{5} x^{2} - \frac{1}{x}

= x^{3} - \frac{3}{5} x^{2} - \frac{1}{x}

Add a constant to the solution

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

\frac{d}{d x} ((cos (x)) (2 x^{3}))

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

\frac{d}{d x} \frac{5 x}{ln ( ( 4 x - 1 ) ^{2} )}

x \to \infty lim (g (x))

5 x^{2} - 3 x - 5

What is the integral of (15x^4-6x^3+5)/(5x^2) ?
The integral of (15x^4-6x^3+5)/(5x^2) is x^3-3/5 x^2-1/x+C