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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (z^2)/(z+1)

\int \frac{z ^{2}}{z + 1} d z

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

Solution steps

\int \frac{z ^{2}}{z + 1} d z

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 = z + 1

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

Add a constant to the solution

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

y^{^{''}} + y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 0

\frac{3}{t}

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

n = 1 \sum \infty \frac{1}{n ^{2} + 9}

\frac{d}{d x} (- \frac{18}{x ^{3}})

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