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Differentiate w.r.t. x
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\int x\left(4+4x^{3}+\left(x^{3}\right)^{2}\right)\mathrm{d}x
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+x^{3}\right)^{2}.
\int x\left(4+4x^{3}+x^{6}\right)\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.
\int 4x+4x^{4}+x^{7}\mathrm{d}x
Use the distributive property to multiply x by 4+4x^{3}+x^{6}.
\int 4x\mathrm{d}x+\int 4x^{4}\mathrm{d}x+\int x^{7}\mathrm{d}x
Integrate the sum term by term.
4\int x\mathrm{d}x+4\int x^{4}\mathrm{d}x+\int x^{7}\mathrm{d}x
Factor out the constant in each of the terms.
2x^{2}+4\int x^{4}\mathrm{d}x+\int x^{7}\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}+\frac{4x^{5}}{5}+\int x^{7}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 4 times \frac{x^{5}}{5}.
2x^{2}+\frac{4x^{5}}{5}+\frac{x^{8}}{8}
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}.
\frac{x^{8}}{8}+\frac{4x^{5}}{5}+2x^{2}
Simplify.
\frac{x^{8}}{8}+\frac{4x^{5}}{5}+2x^{2}+С
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.