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

The integral of (y^2-1/(y^2))^3 is (y^7)/7-y^3-3/y+1/(5y^5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (y^2-1/(y^2))^3

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

\frac{y ^{7}}{7} - y^{3} - \frac{3}{y} + \frac{1}{5 y ^{5}} + C

Solution steps

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

Expand

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

= \int y^{6} - 3 y^{2} + \frac{3}{y ^{2}} - \frac{1}{y ^{6}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{6} d y - \int 3 y^{2} d y + \int \frac{3}{y ^{2}} d y - \int \frac{1}{y ^{6}} d y

\int y^{6} d y = \frac{y ^{7}}{7}

\int 3 y^{2} d y = y^{3}

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

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

= \frac{y ^{7}}{7} - y^{3} - \frac{3}{y} - (- \frac{1}{5 y ^{5}})

Simplify

= \frac{y ^{7}}{7} - y^{3} - \frac{3}{y} + \frac{1}{5 y ^{5}}

Add a constant to the solution

= \frac{y ^{7}}{7} - y^{3} - \frac{3}{y} + \frac{1}{5 y ^{5}} + C

\int cos (a t) d t

\int_{0}^{1} y e^{y} d y

x \to 0 lim (\frac{( x + 3 ) ^{2} - 9}{x})

x (10 - 2 x) (16 - 2 x)

y^{^{'}} + 2 y = e^{x}

What is the integral of (y^2-1/(y^2))^3 ?
The integral of (y^2-1/(y^2))^3 is (y^7)/7-y^3-3/y+1/(5y^5)+C