What is the integral of 1/(9Z^2+4) ?

The integral of 1/(9Z^2+4) is 1/6 arctan(3/2 Z)+C

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integral of 1/(9Z^2+4)

\int \frac{1}{9 Z ^{2} + 4} d Z

\frac{1}{6} arctan (\frac{3}{2} Z) + C

Solution steps

\int \frac{1}{9 Z ^{2} + 4} d Z

Apply Integral Substitution

= \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{6} arctan (u)

Substitute back

u = \frac{3}{2} Z

= \frac{1}{6} arctan (\frac{3}{2} Z)

Add a constant to the solution

= \frac{1}{6} arctan (\frac{3}{2} Z) + C

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What is the integral of 1/(9Z^2+4) ?
The integral of 1/(9Z^2+4) is 1/6 arctan(3/2 Z)+C