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

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

Evaluate

$$\frac{x^{2}}{2}+\frac{8x^{\frac{3}{2}}}{9}+\frac{4x}{9}+С$$

Solution Steps

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

$$\int \left(\sqrt{x}\right)^{2}+\frac{4}{3}\sqrt{x}+\frac{4}{9}\mathrm{d}x$$

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

$$\int x+\frac{4}{3}\sqrt{x}+\frac{4}{9}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int \frac{4\sqrt{x}}{3}\mathrm{d}x+\int \frac{4}{9}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x\mathrm{d}x+\frac{4\int \sqrt{x}\mathrm{d}x}{3}+\int \frac{4}{9}\mathrm{d}x$$

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}.

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

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 \frac{4}{3} times \frac{2x^{\frac{3}{2}}}{3}.

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

Find the integral of \frac{4}{9} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{2}}{2}+\frac{8x^{\frac{3}{2}}}{9}+\frac{4x}{9}$$

Simplify.

$$\frac{4x}{9}+\frac{8x^{\frac{3}{2}}}{9}+\frac{x^{2}}{2}$$

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.

$$\frac{4x}{9}+\frac{8x^{\frac{3}{2}}}{9}+\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$\frac{\left(3\sqrt{x}+2\right)^{2}}{9}$$