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

The integral of ((x^2))/(x+1) is ((x+1)^2)/2-2(x+1)+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{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ + C

Solution steps

\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 ∣

Add a constant to the solution

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

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

\int \frac{1}{2 ( sin ( x ) ) ^{2} + 3 ( cos ( x ) ) ^{2}} d x

\int ey d y

\int x e^{x - 4} d x

\int - x^{2} + x d x

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