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

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

integral

(\frac{1 - x}{y})^{2}

\frac{1}{y ^{2}} (x - x^{2} + \frac{x ^{3}}{3}) + C

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Take the constant out:

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

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

Expand

(1 - x)^{2} : 1 - 2 x + x^{2}

= \frac{1}{y ^{2}} \cdot \int 1 - 2 x + 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

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

\int 1 d x = x

\int 2 x d x = x^{2}

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

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

Add a constant to the solution

= \frac{1}{y ^{2}} (x - x^{2} + \frac{x ^{3}}{3}) + C

\frac{1}{x ^{2} + 2}

(x + 2)^{3}

\frac{( 1 + e ^{- x} )}{x}

\frac{e ^{2 x}}{x}

\frac{1}{( x + 1 ) ^{5}}

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