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

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

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

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

Solution steps

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

Expand

\frac{( 3 - x ) ( 5 + x )}{x ^{4}} : \frac{15}{x ^{4}} - \frac{2}{x ^{3}} - \frac{1}{x ^{2}}

= \int \frac{15}{x ^{4}} - \frac{2}{x ^{3}} - \frac{1}{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

= \int \frac{15}{x ^{4}} d x - \int \frac{2}{x ^{3}} d x - \int \frac{1}{x ^{2}} d x

\int \frac{15}{x ^{4}} d x = - \frac{5}{x ^{3}}

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

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

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

Simplify

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

Add a constant to the solution

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

f (x) = \frac{e ^{3 x}}{x ^{6}}

\frac{d}{d x} (x^{3} + a x^{2} + b x + c)

\frac{d}{d x} (5 x^{2} \cdot ln (x))

(5 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 4

\int_{0}^{6} (x^{2} - 4) d x

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