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

The integral of (3z)/(x^2+z^2) is 3arctan(x/z)+C

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integral of (3z)/(x^2+z^2)

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

3 arctan (\frac{x}{z}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 3 z \cdot \int \frac{1}{z ( u ^{2} + 1 )} d u

Take the constant out:

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

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

Use the common integral:

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

= 3 z \frac{1}{z} arctan (u)

Substitute back

u = \frac{x}{z}

= 3 z \frac{1}{z} arctan (\frac{x}{z})

Simplify

3 z \frac{1}{z} arctan (\frac{x}{z}) : 3 arctan (\frac{x}{z})

= 3 arctan (\frac{x}{z})

Add a constant to the solution

= 3 arctan (\frac{x}{z}) + C

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What is the integral of (3z)/(x^2+z^2) ?
The integral of (3z)/(x^2+z^2) is 3arctan(x/z)+C