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

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

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

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

Solution steps

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

Expand

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

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

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

\int 3 d x = 3 x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

g (t) = 4 cos (4 t) - 9 sin (4 t) + 2 cos (10 t)

\frac{\partial}{\partial x} (r sin (t) sin (x))

\frac{\partial}{\partial x} (e^{- (x^{2} + y^{2} + 2 z^{2})})

\int \frac{10 + u}{u} d u

\int \frac{3}{x} + \frac{x x}{4} + C d x

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