Solve ∫ (x+1/2)*(x^2+x)^2 wrt x

$$\int ( x + \frac { 1 } { 2 } ) \cdot ( x ^ { 2 } + x ) ^ { 2 } d x$$

Evaluate

$$\frac{x^{6}}{6}+\frac{x^{5}}{2}+\frac{x^{4}}{2}+\frac{x^{3}}{6}+С$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int \left(x+\frac{1}{2}\right)\left(x^{4}+2x^{2}x+x^{2}\right)\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int \left(x+\frac{1}{2}\right)\left(x^{4}+2x^{3}+x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x+\frac{1}{2} by x^{4}+2x^{3}+x^{2} and combine like terms.

$$\int x^{5}+\frac{5}{2}x^{4}+2x^{3}+\frac{1}{2}x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{5}\mathrm{d}x+\int \frac{5x^{4}}{2}\mathrm{d}x+\int 2x^{3}\mathrm{d}x+\int \frac{x^{2}}{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{5}\mathrm{d}x+\frac{5\int x^{4}\mathrm{d}x}{2}+2\int x^{3}\mathrm{d}x+\frac{\int x^{2}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+\frac{5\int x^{4}\mathrm{d}x}{2}+2\int x^{3}\mathrm{d}x+\frac{\int x^{2}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply \frac{5}{2} times \frac{x^{5}}{5}.

$$\frac{x^{6}}{6}+\frac{x^{5}}{2}+2\int x^{3}\mathrm{d}x+\frac{\int x^{2}\mathrm{d}x}{2}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{6}+\frac{x^{5}}{2}+\frac{x^{4}}{2}+\frac{\int x^{2}\mathrm{d}x}{2}$$

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

$$\frac{x^{6}}{6}+\frac{x^{5}}{2}+\frac{x^{4}}{2}+\frac{x^{3}}{6}$$

Simplify.

$$\frac{x^{4}}{2}+\frac{x^{5}}{2}+\frac{x^{6}}{6}+\frac{x^{3}}{6}$$

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.

$$\frac{x^{4}}{2}+\frac{x^{5}}{2}+\frac{x^{6}}{6}+\frac{x^{3}}{6}+С$$

Differentiate w.r.t. x

$$\frac{\left(2x+1\right)\left(x\left(x+1\right)\right)^{2}}{2}$$