3. $int_{1}^{3}(3x^{2}+\frac{4}{x^{2}})dx$

Answer:

$26$

Explanation:

Step1: Split the integral

$$\int_{1}^{3}(3x^{2}+\frac{4}{x^{2}})dx=\int_{1}^{3}3x^{2}dx+\int_{1}^{3}\frac{4}{x^{2}}dx$$

Step2: Rewrite the second - integral

$$\int_{1}^{3}\frac{4}{x^{2}}dx = 4\int_{1}^{3}x^{- 2}dx$$

Step3: Integrate term - by - term

For $\int_{1}^{3}3x^{2}dx$, using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $3\times\frac{x^{3}}{3}\big|{1}^{3}=x^{3}\big|{1}^{3}=3^{3}-1^{3}=27 - 1 = 26$. For $4\int_{1}^{3}x^{-2}dx$, using the power rule, we have $4\times\frac{x^{-2 + 1}}{-2+1}\big|{1}^{3}=4\times(-x^{-1})\big|{1}^{3}=-4(\frac{1}{3}-\frac{1}{1})=-4\times(-\frac{2}{3})=\frac{8}{3}$. The sum is $26+\frac{8}{3}-\frac{8}{3}=26$.