How to integrate $\int \frac{z}{x^2 + z^2} dx$?

Question

How to integrate $\int \frac{z}{x^2 + z^2} dx$?

Any help is appreciated!

Thank you!

Answer 1

First, notice that since the integral you gave is with respect to $x$, we treat $z$ as a constant. Then using the following antiderivative formula: $\displaystyle \int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\frac{x}{a}+C$, we find that $$\int\frac{z}{x^2+z^2}\,dx=z\int\frac{1}{x^2+z^2}\,dx=z\cdot\frac{1}{z}\arctan\frac{x}{z}+C=\arctan\frac{x}{z}+C.$$

Answer 2

Put x = z tanu

dx = z sec^2u du

$\int \frac{z}{z^2\tan^2u + z^2} z \sec^2u du$

$\int \frac{1}{\tan^2u + 1} \sec^2u du$

$\int \frac{1}{\sec^2u} \sec^2u du$

$\int du$

= u

= $\tan^{-1}\frac{x}{z}$