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

The integral of xysqrt(x^2+y^2) is (x(x^2+y^2)^{3/2})/3+C

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

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

\frac{x ( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{2} d u

= x \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= x \frac{u ^{3}}{3}

Substitute back

u = x^{2} + y^{2}

= x \frac{( x ^{2} + y ^{2} ) ^{3}}{3}

Simplify

x \frac{( x ^{2} + y ^{2} ) ^{3}}{3} : \frac{x ( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3}

= \frac{x ( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3}

Add a constant to the solution

= \frac{x ( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3} + C

y = x + \frac{2}{x}

\int_{0}^{4} \frac{x}{2 x + 1} d x

x \to 0 + lim (\frac{5}{x + x ^{2}})

\frac{f ( a + h ) - f ( a - h )}{2 h}

x \to - 1 lim (∣ x - 1 ∣)

What is the integral of xysqrt(x^2+y^2) ?
The integral of xysqrt(x^2+y^2) is (x(x^2+y^2)^{3/2})/3+C