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

The integral of ((u-1)^2)/u is (u^2)/2-2u+ln|u|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((u-1)^2)/u

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

\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣ + C

Solution steps

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

Expand

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

= \int u - 2 + \frac{1}{u} d u

Apply the Sum Rule:

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

= \int u d u - \int 2 d u + \int \frac{1}{u} d u

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Add a constant to the solution

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣ + C

f (x) = 10 x

\frac{d}{d x} (x^{3} + 3 x - 6)

\int \frac{7 - 98 x}{36 - 49 x ^{2}} d x

x = y^{2}, x = y^{3}

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

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