evaluate the definite integral below. ∫₂³ (16x + 12) / ∛(2x² + 3x + 8) dx

Explanation:

Step1: Use substitution

Let $u = 2x^{2}+3x + 8$. Then $du=(4x + 3)dx$. Notice that $16x+12 = 4(4x + 3)$. So, when $x = 2$, $u=2\times2^{2}+3\times2 + 8=8 + 6+8=22$; when $x = 3$, $u=2\times3^{2}+3\times3 + 8=18+9 + 8=35$.

Step2: Rewrite the integral

The integral $\int_{2}^{3}\frac{16x + 12}{\sqrt[3]{2x^{2}+3x + 8}}dx=4\int_{22}^{35}u^{-\frac{1}{3}}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $4\int_{22}^{35}u^{-\frac{1}{3}}du=4\times\frac{u^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\big|_{22}^{35}$.

Step4: Simplify the antiderivative

$4\times\frac{u^{\frac{2}{3}}}{\frac{2}{3}}\big|{22}^{35}=6u^{\frac{2}{3}}\big|{22}^{35}$.

Step5: Evaluate the definite integral

$6\times(35^{\frac{2}{3}}-22^{\frac{2}{3}})=6( \sqrt[3]{35^{2}}-\sqrt[3]{22^{2}})=6(\sqrt[3]{1225}-\sqrt[3]{484})$.

Answer:

$6(\sqrt[3]{1225}-\sqrt[3]{484})$