What is the integral of xysqrt(1+x^2+y^2) ?

The integral of xysqrt(1+x^2+y^2) is (x(1+x^2+y^2)^{3/2})/3+C

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integral of xysqrt(1+x^2+y^2)

\int x y 1 + x^{2} + y^{2} d y

\frac{x ( 1 + x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3} + C

Solution steps

\int x y 1 + 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 1 + x^{2} + y^{2} y d y

Apply u-substitution

: \int \frac{u}{2} d u

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

Take the constant out:

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

= x \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= x \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + x^{2} + y^{2}

= x \frac{1}{2} \cdot \frac{2}{3} (1 + x^{2} + y^{2})^{\frac{3}{2}}

Simplify

x \frac{1}{2} \cdot \frac{2}{3} (1 + x^{2} + y^{2})^{\frac{3}{2}} : \frac{x ( 1 + x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3}

= \frac{x ( 1 + x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3}

Add a constant to the solution

= \frac{x ( 1 + x ^{2} + y ^{2} ) ^{\frac{3}{2}}}{3} + C

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What is the integral of xysqrt(1+x^2+y^2) ?
The integral of xysqrt(1+x^2+y^2) is (x(1+x^2+y^2)^{3/2})/3+C