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

The integral of (z^2)/(z^2+a^2) is -aarctan(z/a)+z+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^2+a^2)

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

- a arctan (\frac{z}{a}) + z + C

Solution steps

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

Apply Integral Substitution

= \int \frac{a u ^{2}}{u ^{2} + 1} d u

Take the constant out:

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

= a \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= a \cdot \int - \frac{1}{u ^{2} + 1} + 1 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

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

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

\int 1 d u = u

= a (- arctan (u) + u)

Substitute back

u = \frac{z}{a}

= a (- arctan (\frac{z}{a}) + \frac{z}{a})

Simplify

a (- arctan (\frac{z}{a}) + \frac{z}{a}) : - a arctan (\frac{z}{a}) + z

= - a arctan (\frac{z}{a}) + z

Add a constant to the solution

= - a arctan (\frac{z}{a}) + z + C

\int \frac{4 sin ( x )}{( 1 + cos ^{2} ( x ) )} d x

\int \frac{2 x ^{2} - 4 x + 3}{x - 2} d x

\int - π sin (π) θ d θ

\int x^{4} (1 - x)^{4} d x

\int 3 x^{4} - 17 + \frac{1}{x} - sin (x) d x

What is the integral of (z^2)/(z^2+a^2) ?
The integral of (z^2)/(z^2+a^2) is -aarctan(z/a)+z+C