Question

Solve ${\int }_{\frac{1}{e}}^{e^2}\left(\frac{x^3+1}{x}\right)dx$

• A. $\frac{e^6-1}{6e^2}$
• B. $\frac{e^{12}+2e^6-1}{6}$
• C. $\frac{e^9+e^3-1}{e^3}$
• D. $\frac{e^9+9e^3-1}{3e^3}$
• E. $\frac{e^9-1}{e^3}+\ln 2$

Answer 1

The correct option is D $\frac{e^9+9e^3-1}{3e^3}$

The value of ${\int }_{1/e}^{e^2}\frac{x^3+1}{x}dx$

$={\int }_{1/e}^{e^2}(x^2+\frac{1}{x})dx$

$={\int }_{1/e}^{e^2}{x}^{2}dx+{\int }_{1/e}^{e^2}\frac{1}{x}dx$

$=\frac{1}{3}(e^6-\frac{1}{e^3})+(2+1)$

$=\frac{e^9+9e^3-1}{3e^3}$