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

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

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integral of (x^2+y^2)^{1/2}

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

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

Solution steps

\int (x^{2} + y^{2})^{\frac{1}{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{1}{2}} + \frac{y ^{2}}{2} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣}

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

Add a constant to the solution

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

\int - x e^{- x} d x

\int (4 x + 7) d x

x \to 3 lim (8)

\frac{d}{d x} (3 x \cdot sin (x))

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