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Differentiate w.r.t. x
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\int \left(\left(x^{8}\right)^{3}+18\left(x^{8}\right)^{2}+108x^{8}+216\right)\times 8x^{7}\mathrm{d}x
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x^{8}+6\right)^{3}.
\int \left(x^{24}+18\left(x^{8}\right)^{2}+108x^{8}+216\right)\times 8x^{7}\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 8 and 3 to get 24.
\int \left(x^{24}+18x^{16}+108x^{8}+216\right)\times 8x^{7}\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 8 and 2 to get 16.
\int \left(8x^{24}+144x^{16}+864x^{8}+1728\right)x^{7}\mathrm{d}x
Use the distributive property to multiply x^{24}+18x^{16}+108x^{8}+216 by 8.
\int 8x^{31}+144x^{23}+864x^{15}+1728x^{7}\mathrm{d}x
Use the distributive property to multiply 8x^{24}+144x^{16}+864x^{8}+1728 by x^{7}.
\int 8x^{31}\mathrm{d}x+\int 144x^{23}\mathrm{d}x+\int 864x^{15}\mathrm{d}x+\int 1728x^{7}\mathrm{d}x
Integrate the sum term by term.
8\int x^{31}\mathrm{d}x+144\int x^{23}\mathrm{d}x+864\int x^{15}\mathrm{d}x+1728\int x^{7}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{32}}{4}+144\int x^{23}\mathrm{d}x+864\int x^{15}\mathrm{d}x+1728\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^{31}\mathrm{d}x with \frac{x^{32}}{32}. Multiply 8 times \frac{x^{32}}{32}.
\frac{x^{32}}{4}+6x^{24}+864\int x^{15}\mathrm{d}x+1728\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^{23}\mathrm{d}x with \frac{x^{24}}{24}. Multiply 144 times \frac{x^{24}}{24}.
\frac{x^{32}}{4}+6x^{24}+54x^{16}+1728\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^{15}\mathrm{d}x with \frac{x^{16}}{16}. Multiply 864 times \frac{x^{16}}{16}.
\frac{x^{32}}{4}+6x^{24}+54x^{16}+216x^{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}. Multiply 1728 times \frac{x^{8}}{8}.
216x^{8}+54x^{16}+6x^{24}+\frac{x^{32}}{4}+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.