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

= \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

\frac{9}{s ^{2} + 9 s + 9} \cdot \frac{1}{s}

f (x) = x^{4} + 9 x^{2} - x, at x = 1

n = 3 \sum \infty (\frac{4}{7})^{n}

\int \frac{1}{cos ^{2} ( 4 x )} d x

\frac{d}{d x} (e^{x} - 4)

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