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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^2+1)/(x^3))^2

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

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

Solution steps

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

Simplify

(\frac{x ^{2} + 1}{x ^{3}})^{2} : \frac{( x ^{2} + 1 ) ^{2}}{x ^{6}}

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

Expand

\frac{( x ^{2} + 1 ) ^{2}}{x ^{6}} : \frac{1}{x ^{2}} + \frac{2}{x ^{4}} + \frac{1}{x ^{6}}

= \int \frac{1}{x ^{2}} + \frac{2}{x ^{4}} + \frac{1}{x ^{6}} 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 ^{2}} d x + \int \frac{2}{x ^{4}} d x + \int \frac{1}{x ^{6}} d x

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

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

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

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

Add a constant to the solution

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

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

\int 1 1^{4 x} d x

\int \frac{2 x}{( x - 4 ) ( x + 3 )} d x

\int t^{3} (25 + 81 t^{2})^{\frac{5}{2}} d t

\int 9 t^{2} sin (t^{3}) d t

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