What is the integral of (z^2)/(z^3+1) ?

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

\int \frac{z ^{2}}{z ^{3} + 1} d z

\frac{1}{3} ln z^{3} + 1 + C

Solution steps

\int \frac{z ^{2}}{z ^{3} + 1} d z

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = z^{3} + 1

= \frac{1}{3} ln z^{3} + 1

Add a constant to the solution

= \frac{1}{3} ln z^{3} + 1 + C

- 9 (5 x^{2} + 4)^{- 6}

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

\int e^{t a n (x)} sec^{2} (x) d x

y = 4 x^{2}

f (x) = ln (7)

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