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

The integral of 1/(sqrt(x^2+z^2)) is ln((|x+sqrt(z^2+x^2)|)/(|z|))+C

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

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

ln (\frac{x + z ^{2} + x ^{2}}{∣ z ∣}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

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

= ln tan (arctan (\frac{1}{z} x)) + sec (arctan (\frac{1}{z} x))

Simplify

ln tan (arctan (\frac{1}{z} x)) + sec (arctan (\frac{1}{z} x)) : ln (\frac{x + z ^{2} + x ^{2}}{∣ z ∣})

= ln (\frac{x + z ^{2} + x ^{2}}{∣ z ∣})

Add a constant to the solution

= ln (\frac{x + z ^{2} + x ^{2}}{∣ z ∣}) + C

f (x) = 4 x + 1

\frac{d y}{d t} - (\frac{3}{t}) y = 2 t^{3} e^{2 t}

\frac{\partial}{\partial x} (cos (5 x - 2 y^{2}))

x^{2} + 8

n = 1 \sum \infty \frac{1}{4 n ^{6} + n + 1}

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