$\displaystyle \int \limits_1^e \left( \frac{x^2-1}{x} \right)   dx =$

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Break apart the fraction in order to use the power rule and rule for natural logarithms.

$$\int \limits_1^e \left( \frac{x^2-1}{x} \right)   dx = \int \limits_1^e \left( \frac{x^2}{x}-\frac{1}{x} \right)   dx$$ $$= \int \limits_1^e \left( x-\frac{1}{x} \right)   dx$$ $$= \left[ \frac{1}{2}x^2-\ln |x| \right]_1^e$$ $$= \frac{1}{2}(e)^2-\ln|e|-\left(\frac{1}{2}(1)^2-\ln|1|\right)$$ $$= \frac{1}{2}e^2-1-\left(\frac{1}{2}-0\right)$$ $$= \frac{1}{2}e^2-1-\frac{1}{2}$$ $$=\boxed{\frac{e^2}{2}-\frac{3}{2}}$$