Solve ∫ frac{(x+2sqrt{x})(x-2sqrt{x})}{x}

$$\int \frac { ( x + 2 \sqrt { x } ) ( x - 2 \sqrt { x } ) } { x }$$

Evaluate

$$\frac{x^{2}}{2}-4x+С$$

Solution Steps

Consider \left(x+2\sqrt{x}\right)\left(x-2\sqrt{x}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\int \frac{x^{2}-\left(2\sqrt{x}\right)^{2}}{x}\mathrm{d}x$$

Expand \left(2\sqrt{x}\right)^{2}.

$$\int \frac{x^{2}-2^{2}\left(\sqrt{x}\right)^{2}}{x}\mathrm{d}x$$

Calculate 2 to the power of 2 and get 4.

$$\int \frac{x^{2}-4\left(\sqrt{x}\right)^{2}}{x}\mathrm{d}x$$

Calculate \sqrt{x} to the power of 2 and get x.

$$\int \frac{x^{2}-4x}{x}\mathrm{d}x$$

Factor the expressions that are not already factored in \frac{x^{2}-4x}{x}.

$$\int \frac{x\left(x-4\right)}{x}\mathrm{d}x$$

Cancel out x in both numerator and denominator.

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

Integrate the sum term by term.

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

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}.

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}-4x$$

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.

$$\frac{x^{2}}{2}-4x+С$$

Differentiate w.r.t. x

$$x-4$$