Solve ∫ {left(3-2sqrt{x}right)}^2 wrt x

$$\int{ { \left(3-2 \sqrt{ x } \right) }^{ 2 } }d x$$

Evaluate

$$2x^{2}-8x^{\frac{3}{2}}+9x+С$$

Solution Steps

Multiply -1 and 2 to get -2.

$$\int \left(3-2\sqrt{x}\right)^{2}\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-2\sqrt{x}\right)^{2}.

$$\int 9-12\sqrt{x}+4\left(\sqrt{x}\right)^{2}\mathrm{d}x$$

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

$$\int 9-12\sqrt{x}+4x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9\mathrm{d}x+\int -12\sqrt{x}\mathrm{d}x+\int 4x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}x-12\int \sqrt{x}\mathrm{d}x+4\int x\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-12\int \sqrt{x}\mathrm{d}x+4\int x\mathrm{d}x$$

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify. Multiply -12 times \frac{2x^{\frac{3}{2}}}{3}.

$$9x-8x^{\frac{3}{2}}+4\int x\mathrm{d}x$$

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}.

$$9x-8x^{\frac{3}{2}}+2x^{2}$$

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.

$$2x^{2}-8x^{\frac{3}{2}}+9x+С$$

Differentiate w.r.t. x

$$\left(2\sqrt{x}-3\right)^{2}$$