Question

determine the following indefinite integral. check your work by differentiation.
int \frac{8}{9z^{2}+9} dz
int \frac{8}{9z^{2}+9} dz=square

Explanation:

Step1: Simplify the integrand

First, factor out 9 from the denominator: $\int\frac{8}{9z^{2}+9}dz=\int\frac{8}{9(z^{2} + 1)}dz=\frac{8}{9}\int\frac{1}{z^{2}+1}dz$.

Step2: Recall the integral formula

We know that $\int\frac{1}{x^{2}+1}dx=\arctan(x)+C$. Here $x = z$, so $\frac{8}{9}\int\frac{1}{z^{2}+1}dz=\frac{8}{9}\arctan(z)+C$.

Step3: Check by differentiation

Differentiate $\frac{8}{9}\arctan(z)+C$ using the formula $\frac{d}{dz}\arctan(z)=\frac{1}{z^{2}+1}$. The derivative of $\frac{8}{9}\arctan(z)+C$ is $\frac{8}{9}\times\frac{1}{z^{2}+1}=\frac{8}{9z^{2}+9}$, which is the original integrand.

Answer:

$\frac{8}{9}\arctan(z)+C$