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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

\frac{3 x ^{2} - x + 5}{x ^{4}} : \frac{3}{x ^{2}} - \frac{1}{x ^{3}} + \frac{5}{x ^{4}}

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

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

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

Simplify

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

Add a constant to the solution

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

y^{^{'}} = - \frac{x}{y}

x \to - 3 lim (\frac{x + 2}{x - 2})

\frac{\partial}{\partial x} (cos^{2} (x) y)

\frac{d}{d a} (sec (a) \cdot tan (a) \cdot cos (a) \cdot csc (a))

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

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