∫₃⁵ (2x² + x + 4)/(x - 1) dx = \na (139/9) - 23 \nb \nc 22 + 7(ln4 - ln2) \nd 24 + 8(ln4 - ln2)
∫₃⁵ (2x² + x + 4)/(x - 1) dx = \na (139/9) - 23 \nb \nc 22 + 7(ln4 - ln2) \nd 24 + 8(ln4 - ln2)
Answer
Explanation:
Step1: Perform polynomial long - division
Divide $2x^{2}+x + 4$ by $x - 1$. We have $2x^{2}+x + 4=(x - 1)(2x+3)+7$. So, $\frac{2x^{2}+x + 4}{x - 1}=2x + 3+\frac{7}{x - 1}$.
Step2: Integrate term - by - term
$\int_{3}^{5}(2x + 3+\frac{7}{x - 1})dx=\int_{3}^{5}2xdx+\int_{3}^{5}3dx+\int_{3}^{5}\frac{7}{x - 1}dx$. For $\int_{3}^{5}2xdx=x^{2}\big|{3}^{5}=5^{2}-3^{2}=25 - 9 = 16$. For $\int{3}^{5}3dx=3x\big|{3}^{5}=3(5 - 3)=6$. For $\int{3}^{5}\frac{7}{x - 1}dx=7\ln|x - 1|\big|_{3}^{5}=7(\ln4-\ln2)$.
Step3: Sum up the results
$\int_{3}^{5}(2x + 3+\frac{7}{x - 1})dx=16 + 6+7(\ln4-\ln2)=22+7(\ln4-\ln2)$.
Answer:
C. $22 + 7(\ln4-\ln2)$