Solve ∫ (x+1)^2-(x-1)^2

$$\int ( x + 1 ) ^ { 2 } - ( x - 1 ) ^ { 2 }$$

Evaluate

$$2x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$\int x^{2}+2x+1-\left(x-1\right)^{2}\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\int x^{2}+2x+1-\left(x^{2}-2x+1\right)\mathrm{d}x$$

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$\int x^{2}+2x+1-x^{2}+2x-1\mathrm{d}x$$

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

$$\int 2x+1+2x-1\mathrm{d}x$$

Combine 2x and 2x to get 4x.

$$\int 4x+1-1\mathrm{d}x$$

Subtract 1 from 1 to get 0.

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

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

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

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}. Multiply 4 times \frac{x^{2}}{2}.

$$2x^{2}$$

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.

$$2x^{2}+С$$

Differentiate w.r.t. x

$$4x$$