Solve ∫ x^3(x^4+3)^2 wrt x=

$$\int x ^ { 3 } ( x ^ { 4 } + 3 ) ^ { 2 } d x =$$

Evaluate

$$\frac{x^{12}}{12}+\frac{3x^{8}}{4}+\frac{9x^{4}}{4}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{4}+3\right)^{2}.

$$\int x^{3}\left(\left(x^{4}\right)^{2}+6x^{4}+9\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$\int x^{3}\left(x^{8}+6x^{4}+9\right)\mathrm{d}x$$

Use the distributive property to multiply x^{3} by x^{8}+6x^{4}+9.

$$\int x^{11}+6x^{7}+9x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{11}\mathrm{d}x+\int 6x^{7}\mathrm{d}x+\int 9x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{11}\mathrm{d}x+6\int x^{7}\mathrm{d}x+9\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{11}\mathrm{d}x with \frac{x^{12}}{12}.

$$\frac{x^{12}}{12}+6\int x^{7}\mathrm{d}x+9\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}. Multiply 6 times \frac{x^{8}}{8}.

$$\frac{x^{12}}{12}+\frac{3x^{8}}{4}+9\int x^{3}\mathrm{d}x$$

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

$$\frac{x^{12}}{12}+\frac{3x^{8}}{4}+\frac{9x^{4}}{4}$$

Simplify.

$$\frac{9x^{4}}{4}+\frac{3x^{8}}{4}+\frac{x^{12}}{12}$$

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.

$$\frac{9x^{4}}{4}+\frac{3x^{8}}{4}+\frac{x^{12}}{12}+С$$

Differentiate w.r.t. x

$$x^{3}\left(x^{4}+3\right)^{2}$$