Solve ∫ x(x^2+3)^2 wrt x

$$\int x ( x ^ { 2 } + 3 ) ^ { 2 } d x$$

Evaluate

$$\frac{x^{6}}{6}+\frac{3x^{4}}{2}+\frac{9x^{2}}{2}+С$$

Solution Steps

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

$$\int x\left(\left(x^{2}\right)^{2}+6x^{2}+9\right)\mathrm{d}x$$

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

$$\int x\left(x^{4}+6x^{2}+9\right)\mathrm{d}x$$

Use the distributive property to multiply x by x^{4}+6x^{2}+9.

$$\int x^{5}+6x^{3}+9x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{5}\mathrm{d}x+\int 6x^{3}\mathrm{d}x+\int 9x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{5}\mathrm{d}x+6\int x^{3}\mathrm{d}x+9\int x\mathrm{d}x$$

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

$$\frac{x^{6}}{6}+6\int x^{3}\mathrm{d}x+9\int x\mathrm{d}x$$

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 6 times \frac{x^{4}}{4}.

$$\frac{x^{6}}{6}+\frac{3x^{4}}{2}+9\int x\mathrm{d}x$$

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 9 times \frac{x^{2}}{2}.

$$\frac{x^{6}}{6}+\frac{3x^{4}}{2}+\frac{9x^{2}}{2}$$

Simplify.

$$\frac{9x^{2}}{2}+\frac{3x^{4}}{2}+\frac{x^{6}}{6}$$

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.

$$\frac{9x^{2}}{2}+\frac{3x^{4}}{2}+\frac{x^{6}}{6}+С$$

Differentiate w.r.t. x

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