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

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

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

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

\frac{1}{2} x x^{2} + y^{2} + \frac{y ^{2}}{2} ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣} + C

Solution steps

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

Apply Trigonometric Substitution

: \int y^{2} sec^{3} (u) d u

= \int y^{2} sec^{3} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y^{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= y^{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= y^{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

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

Simplify

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

= \frac{1}{2} x x^{2} + y^{2} + \frac{y ^{2}}{2} ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣}

Add a constant to the solution

= \frac{1}{2} x x^{2} + y^{2} + \frac{y ^{2}}{2} ln \frac{x + y ^{2} + x ^{2}}{∣ y ∣} + C

x \to 5 lim (\frac{1}{x})

\frac{d}{d x} (1 + sin (x))

f (x) = \frac{72}{x}

\frac{d}{d x} ((x + 3)^{2})

4 x^{2} - 5 x

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