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

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

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

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

ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣} + C

Solution steps

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

Apply Trigonometric Substitution

= \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}{y} x)

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

Simplify

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

= ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣}

Add a constant to the solution

= ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣} + C

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