What is the integral of ((1/x-1)^3)/(x^2) ?

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int ln (8) d x

\int - 3 x^{2} + x^{3} d x

\int (2 - 5 x)^{9} d x

\int \frac{99 x - 83}{121 x ^{2} - 176 x + 64} d x

\int cos (4 x) sin (sin (4 x)) d x

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