Solve ∫ x(2+x^2)^2 wrt x

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

Evaluate

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

Solution Steps

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

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

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

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

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

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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}.

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

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.

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

Differentiate w.r.t. x

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