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{d}{d x} (ln (sin (e^{x})))

g (x) = ln (3 x)

9 x^{\frac{1}{3}} + \frac{9}{2} x^{\frac{4}{3}}

\frac{\partial}{\partial x} (- x e^{(x y)})

\int - 2 x + 18 x^{2} - 4 x^{3} + C d x

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