Evaluate the integral \(\int_{1}^{2}\left(x^{2}+\frac{1}{x}\right) d x\).
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Correct Answer: Option B
Explanation:
:
\begin{aligned}
\int_{1}^{2}\left(x^{2}+\frac{1}{x}\right) d x &=\frac{x^{3}}{3}+\ln x \int_{1}^{2} \\
&=\left(\frac{2^{3}}{3}+\ln 2\right)-\left(\frac{1}{3}+\ln 1\right) \\
&=\frac{8}{3}+\ln 2-\frac{1}{3}=\frac{7}{3}+\ln 2
\end{aligned}
:
\begin{aligned}
\int_{1}^{2}\left(x^{2}+\frac{1}{x}\right) d x &=\frac{x^{3}}{3}+\ln x \int_{1}^{2} \\
&=\left(\frac{2^{3}}{3}+\ln 2\right)-\left(\frac{1}{3}+\ln 1\right) \\
&=\frac{8}{3}+\ln 2-\frac{1}{3}=\frac{7}{3}+\ln 2
\end{aligned}