Solve ∫ (sqrt{x}-5)(sqrt{x}+5) wrt x=

$$\int ( \sqrt { x } - 5 ) ( \sqrt { x } + 5 ) d x =$$

Evaluate

$$\frac{x^{2}}{2}-25x+С$$

Solution Steps

Consider \left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\int \left(\sqrt{x}\right)^{2}-5^{2}\mathrm{d}x$$

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

$$\int x-5^{2}\mathrm{d}x$$

Calculate 5 to the power of 2 and get 25.

$$\int x-25\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int -25\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}+\int -25\mathrm{d}x$$

Find the integral of -25 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}-25x$$

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.

$$\frac{x^{2}}{2}-25x+С$$

Differentiate w.r.t. x

$$x-25$$