What is the integral of (xy)/(sqrt(x^2+y^2+1)) ?

The integral of (xy)/(sqrt(x^2+y^2+1)) is ysqrt(1+y^2+x^2)+C

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integral of (xy)/(sqrt(x^2+y^2+1))

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

y 1 + y^{2} + x^{2} + C

Solution steps

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

Take the constant out:

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

= y \cdot \int \frac{x}{1 + y ^{2} + x ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= y \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (3 x^{4} + (2 x - 1)^{2})

\int_{1}^{4} \frac{1}{4 - x} d x

\int \frac{3 x}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{2 ^{x}}{ln ( 2 )} x d x

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

What is the integral of (xy)/(sqrt(x^2+y^2+1)) ?
The integral of (xy)/(sqrt(x^2+y^2+1)) is ysqrt(1+y^2+x^2)+C