Solve ∫ x^3left(x^4+3right)^3 wrt x

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

Evaluate

$$\frac{x^{16}}{16}+\frac{3x^{12}}{4}+\frac{27x^{8}}{8}+\frac{27x^{4}}{4}+С$$

Solution Steps

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

$$\int x^{3}\left(\left(x^{4}\right)^{3}+9\left(x^{4}\right)^{2}+27x^{4}+27\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 4 and 3 to get 12.

$$\int x^{3}\left(x^{12}+9\left(x^{4}\right)^{2}+27x^{4}+27\right)\mathrm{d}x$$

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

$$\int x^{3}\left(x^{12}+9x^{8}+27x^{4}+27\right)\mathrm{d}x$$

Use the distributive property to multiply x^{3} by x^{12}+9x^{8}+27x^{4}+27.

$$\int x^{15}+9x^{11}+27x^{7}+27x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{15}\mathrm{d}x+\int 9x^{11}\mathrm{d}x+\int 27x^{7}\mathrm{d}x+\int 27x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{15}\mathrm{d}x+9\int x^{11}\mathrm{d}x+27\int x^{7}\mathrm{d}x+27\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{15}\mathrm{d}x with \frac{x^{16}}{16}.

$$\frac{x^{16}}{16}+9\int x^{11}\mathrm{d}x+27\int x^{7}\mathrm{d}x+27\int x^{3}\mathrm{d}x$$

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}. Multiply 9 times \frac{x^{12}}{12}.

$$\frac{x^{16}}{16}+\frac{3x^{12}}{4}+27\int x^{7}\mathrm{d}x+27\int x^{3}\mathrm{d}x$$

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 27 times \frac{x^{8}}{8}.

$$\frac{x^{16}}{16}+\frac{3x^{12}}{4}+\frac{27x^{8}}{8}+27\int x^{3}\mathrm{d}x$$

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

$$\frac{x^{16}}{16}+\frac{3x^{12}}{4}+\frac{27x^{8}}{8}+\frac{27x^{4}}{4}$$

Simplify.

$$\frac{27x^{4}}{4}+\frac{27x^{8}}{8}+\frac{3x^{12}}{4}+\frac{x^{16}}{16}$$

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.

$$\frac{27x^{4}}{4}+\frac{27x^{8}}{8}+\frac{3x^{12}}{4}+\frac{x^{16}}{16}+С$$

Differentiate w.r.t. x

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