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

The integral of ((x+4)(x-2))/(2x^4) is (-3x^2-3x+8)/(6x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int \frac{( x + 4 ) ( x - 2 )}{2 x ^{4}} d x

\frac{- 3 x ^{2} - 3 x + 8}{6 x ^{3}} + C

Solution steps

\int \frac{( x + 4 ) ( x - 2 )}{2 x ^{4}} d x

Take the constant out:

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

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

Apply Integration By Parts

= \frac{1}{2} (\frac{( x + 4 ) ( x - 2 )}{x ^{3}} - \int \frac{2 ( - x ^{2} - 3 x + 16 )}{x ^{4}} d x)

\int \frac{2 ( - x ^{2} - 3 x + 16 )}{x ^{4}} d x = \frac{2}{x} + \frac{3}{x ^{2}} - \frac{32}{3 x ^{3}}

= \frac{1}{2} (\frac{( x + 4 ) ( x - 2 )}{x ^{3}} - (\frac{2}{x} + \frac{3}{x ^{2}} - \frac{32}{3 x ^{3}}))

Simplify

\frac{1}{2} (\frac{( x + 4 ) ( x - 2 )}{x ^{3}} - (\frac{2}{x} + \frac{3}{x ^{2}} - \frac{32}{3 x ^{3}})) : \frac{- 3 x ^{2} - 3 x + 8}{6 x ^{3}}

= \frac{- 3 x ^{2} - 3 x + 8}{6 x ^{3}}

Add a constant to the solution

= \frac{- 3 x ^{2} - 3 x + 8}{6 x ^{3}} + C

What is the integral of ((x+4)(x-2))/(2x^4) ?
The integral of ((x+4)(x-2))/(2x^4) is (-3x^2-3x+8)/(6x^3)+C