What is the integral of 1/(x^2+z^2) ?

The integral of 1/(x^2+z^2) is 1/z arctan(x/z)+C

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integral of 1/(x^2+z^2)

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

\frac{1}{z} arctan (\frac{x}{z}) + C

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

u = \frac{x}{z}

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

Add a constant to the solution

= \frac{1}{z} arctan (\frac{x}{z}) + C

\int \frac{9 e ^{x}}{e ^{x} - 1} d x

\frac{\partial}{\partial x} (x^{4} + 5 x^{2} y)

y^{^{''}} - 7 y^{^{'}} + 10 y = 0

\int \frac{4}{x ( ln ( x ) ) ^{2}} d x

x^{4} - 3 x^{2} + 9

What is the integral of 1/(x^2+z^2) ?
The integral of 1/(x^2+z^2) is 1/z arctan(x/z)+C