Solve ∫ x*(2x+1)^2 wrt x

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

Evaluate

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

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

Use the distributive property to multiply x by 4x^{2}+4x+1.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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}.

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

Simplify.

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

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.

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

Differentiate w.r.t. x

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