show instructions question 4 ∫₁ᵉ (x² - 1)/x dx = o a (e²/2) - (3/2) o b e - (1/e) o c e² - e + (1/2) o d e²…

Explanation:

Step1: Rewrite the integrand

Rewrite $\frac{x^{2}-1}{x}$ as $x-\frac{1}{x}$. So the integral becomes $\int_{1}^{e}(x - \frac{1}{x})dx$.

Step2: Integrate term - by - term

Using the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$) and $\int\frac{1}{x}dx=\ln|x|+C$, we have $\int(x-\frac{1}{x})dx=\frac{x^{2}}{2}-\ln x + C$.

Step3: Apply the fundamental theorem of calculus

Evaluate $\left[\frac{x^{2}}{2}-\ln x\right]_{1}^{e}$. Substitute the upper - limit $x = e$ and lower - limit $x = 1$: $(\frac{e^{2}}{2}-\ln e)-(\frac{1^{2}}{2}-\ln1)$. Since $\ln e = 1$ and $\ln1=0$, we get $\frac{e^{2}}{2}-1-\frac{1}{2}+0=\frac{e^{2}}{2}-\frac{3}{2}$.

Answer:

A. $\frac{e^{2}}{2}-\frac{3}{2}$