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

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\int (7 x^{6} + 9) x^{7} + 9 x d 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