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\int _{1}^{2}x^{2}\left(\left(x^{3}\right)^{2}+2x^{3}+1\right)\mathrm{d}x
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{3}+1\right)^{2}.
\int _{1}^{2}x^{2}\left(x^{6}+2x^{3}+1\right)\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.
\int _{1}^{2}x^{8}+2x^{5}+x^{2}\mathrm{d}x
Use the distributive property to multiply x^{2} by x^{6}+2x^{3}+1.
\int x^{8}+2x^{5}+x^{2}\mathrm{d}x
Evaluate the indefinite integral first.
\int x^{8}\mathrm{d}x+\int 2x^{5}\mathrm{d}x+\int x^{2}\mathrm{d}x
Integrate the sum term by term.
\int x^{8}\mathrm{d}x+2\int x^{5}\mathrm{d}x+\int x^{2}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{9}}{9}+2\int x^{5}\mathrm{d}x+\int x^{2}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}.
\frac{x^{9}}{9}+\frac{x^{6}}{3}+\int x^{2}\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}. Multiply 2 times \frac{x^{6}}{6}.
\frac{x^{9}}{9}+\frac{x^{6}}{3}+\frac{x^{3}}{3}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.
\frac{2^{3}}{3}+\frac{2^{6}}{3}+\frac{2^{9}}{9}-\left(\frac{1^{3}}{3}+\frac{1^{6}}{3}+\frac{1^{9}}{9}\right)
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
\frac{721}{9}
Simplify.