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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= - \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 ∣)

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{( 3 x ^{2} + 6 x + 2 )}{x ^{2} + 3 x + 2} d x

\int \frac{( y ^{3} + 4 y )}{9 + y ^{4}} d y

\int \frac{cos ( θ )}{4 - sin ^{2} ( θ )} d θ

\int 6 x - 9 x^{2} + C d x

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