Solve ∫ (x+y)^2-(x-y)^2]div2xy

$$\int ( x + y ) ^ { 2 } - ( x - y ) ^ { 2 } ] \div 2 x y$$

Evaluate

$$2yx^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+y\right)^{2}.

$$\int x^{2}+2xy+y^{2}-\left(x-y\right)^{2}\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-y\right)^{2}.

$$\int x^{2}+2xy+y^{2}-\left(x^{2}-2xy+y^{2}\right)\mathrm{d}x$$

To find the opposite of x^{2}-2xy+y^{2}, find the opposite of each term.

$$\int x^{2}+2xy+y^{2}-x^{2}+2xy-y^{2}\mathrm{d}x$$

Combine x^{2} and -x^{2} to get 0.

$$\int 2xy+y^{2}+2xy-y^{2}\mathrm{d}x$$

Combine 2xy and 2xy to get 4xy.

$$\int 4xy+y^{2}-y^{2}\mathrm{d}x$$

Combine y^{2} and -y^{2} to get 0.

$$\int 4xy\mathrm{d}x$$

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$4y\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

$$4y\times \frac{x^{2}}{2}$$

Simplify.

$$2yx^{2}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$2yx^{2}+С$$

Differentiate w.r.t. x

$$4xy$$