What is the integral from 0 to z of z/(x^2+z^2) ?

The integral from 0 to z of z/(x^2+z^2) is pi/4

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integral from 0 to z of z/(x^2+z^2)

\int_{0}^{z} \frac{z}{x ^{2} + z ^{2}} d x

\frac{π}{4}

Solution steps

\int_{0}^{z} \frac{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

= z \cdot \int_{0}^{z} \frac{1}{z ^{2} + x ^{2}} d x

Apply Integral Substitution

= z \cdot \int_{0}^{1} \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

= z \frac{1}{z} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= z \frac{1}{z} [arctan (u)]_{0}^{1}

Simplify

z \frac{1}{z} [arctan (u)]_{0}^{1} : [arctan (u)]_{0}^{1}

= [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

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What is the integral from 0 to z of z/(x^2+z^2) ?
The integral from 0 to z of z/(x^2+z^2) is pi/4